fizikzauglom

GRADE 10 PRACTICE

Physics Problems: Electrostatics, Grade 10

Charge, field, potential and capacitors.

30 problems across 6 sub-topics
Practise in the interactive mode

Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.

Problem statements

The answer options are there to check against, and the walkthrough opens under each problem.

30 problems
Position in this set: 1 from 30

Problem #94

Electrostatics: charge and field

An external action transferred from an initially neutral metal sphere to another identical neutral sphere 5\cdot10^{10} electrons. Find the modulus of the charge of each ball and the modulus of the force of their interaction at a distance 40\,\text{см}?

Answer options

  1. \left(8 \cdot 10^{-9}\,\text{Кл};\ 3{,}6 \cdot 10^{-6}\,\text{Н}\right)
  2. \left(4 \cdot 10^{-9}\,\text{Кл};\ 1{,}8 \cdot 10^{-6}\,\text{Н}\right)
  3. \left(1{,}6 \cdot 10^{-8}\,\text{Кл};\ 7{,}2 \cdot 10^{-6}\,\text{Н}\right)
  4. \left(8 \cdot 10^{-8}\,\text{Кл};\ 3{,}6 \cdot 10^{-5}\,\text{Н}\right)
Show the walkthroughHide the walkthrough
Given
n
5\cdot 10^{10}
r
0{,}4\,\text{м}
e
1{,}6\cdot 10^{-19}\,\text{Кл}
Find

qF

  1. \left|q_1\right| = \left|q_2\right| = ne\left|q_1\right| = 5\cdot 10^{10}\cdot 1{,}6\cdot 10^{-19}q = 8\cdot 10^{-9}\,\text{Кл}

    One ball gained extra electrons, the other lost exactly the same number.

  2. F = \dfrac{k\left|q_1\right|\left|q_2\right|}{r^{2}}F = \dfrac{9\cdot 10^{9}\cdot 8\cdot 10^{-9}\cdot 8\cdot 10^{-9}}{0{,}4^{2}}F = 3{,}6\cdot 10^{-6}\,\text{Н}

Answerq = 8\cdot 10^{-9}\,\text{Кл}F = 3{,}6\cdot 10^{-6}\,\text{Н}

Position in this set: 2 from 30

Problem #95

Electrostatics: charge and field

With what force is an electron attracted to a proton in a hydrogen atom at a distance 5\cdot10^{-11}\,\text{м}?

Answer options

  1. 9{,}2 \cdot 10^{-7}\,\text{Н}
  2. 9{,}2 \cdot 10^{-8}\,\text{Н}
  3. 4{,}6 \cdot 10^{-8}\,\text{Н}
  4. 1{,}84 \cdot 10^{-7}\,\text{Н}
Show the walkthroughHide the walkthrough
Given
\left|q_1\right| = \left|q_2\right|
e = 1{,}6\cdot 10^{-19}\,\text{Кл}
r
5\cdot 10^{-11}\,\text{м}
Find

F

  1. F = \dfrac{k e^{2}}{r^{2}}F = \dfrac{9\cdot 10^{9}\cdot\left(1{,}6\cdot 10^{-19}\right)^{2}}{\left(5\cdot 10^{-11}\right)^{2}}F = 9{,}2\cdot 10^{-8}\,\text{Н}

AnswerF = 9{,}2\cdot 10^{-8}\,\text{Н}

Position in this set: 3 from 30

Problem #96

Electrostatics: charge and field

Determine the field strength at a distance 60\,\text{см} from charge 2\cdot10^{-8}\,\text{Кл} and the force acting on the electron there.

Answer options

  1. \left(1000\,\text{Н/Кл};\ 1{,}6 \cdot 10^{-16}\,\text{Н}\right)
  2. \left(5000\,\text{Н/Кл};\ 8 \cdot 10^{-16}\,\text{Н}\right)
  3. \left(500\,\text{Н/Кл};\ 8 \cdot 10^{-17}\,\text{Н}\right)
  4. \left(250\,\text{Н/Кл};\ 4 \cdot 10^{-17}\,\text{Н}\right)
Show the walkthroughHide the walkthrough
Given
r
60\,\text{см}0{,}6\,\text{м}
q
2\cdot 10^{-8}\,\text{Кл}
\left|q_0\right|
e = 1{,}6\cdot 10^{-19}\,\text{Кл}
Find

EF

  1. E = \dfrac{k\left|q\right|}{r^{2}}E = \dfrac{9\cdot 10^{9}\cdot 2\cdot 10^{-8}}{0{,}6^{2}}E = 500\,\tfrac{\text{Н}}{\text{Кл}} = 500\,\tfrac{\text{В}}{\text{м}}
  2. F = \left|q_0\right|EF = 1{,}6\cdot 10^{-19}\cdot 500F = 8\cdot 10^{-17}\,\text{Н}

AnswerE = 500\,\tfrac{\text{Н}}{\text{Кл}}F = 8\cdot 10^{-17}\,\text{Н}

Position in this set: 4 from 30

Problem #97

Electrostatics: charge and field

Metal charged ball with radius 20\,\text{см} has a surface charge density 4\cdot10^{-7}\,\text{Кл/м}^{2}. Find the field strength inside the sphere, immediately outside its surface, and at a distance of 10\,\text{см} from the surface.

Answer options

  1. \left(0\,\text{Н/Кл};\ 22500\,\text{Н/Кл};\ 10000\,\text{Н/Кл}\right)
  2. \left(0\,\text{Н/Кл};\ 90000\,\text{Н/Кл};\ 40000\,\text{Н/Кл}\right)
  3. \left(0\,\text{Н/Кл};\ 4{,}5 \cdot 10^{5}\,\text{Н/Кл};\ 2 \cdot 10^{5}\,\text{Н/Кл}\right)
  4. \left(0\,\text{Н/Кл};\ 45000\,\text{Н/Кл};\ 20000\,\text{Н/Кл}\right)
Show the walkthroughHide the walkthrough
Given
R
20\,\text{см}0{,}2\,\text{м}
\sigma
4\cdot 10^{-7}\,\tfrac{\text{Кл}}{\text{м}^{2}}
S
0{,}1\,\text{м}
Find

E_1E_2E_3

  1. E_1 = 0

    There is no field inside a charged conductor.

  2. q = \sigma\cdot 4\pi R^{2}q = 4\cdot 10^{-7}\cdot 4\cdot 3{,}14\cdot 0{,}2^{2}q = 2\cdot 10^{-7}\,\text{Кл}
  3. E_2 = \dfrac{kq}{R^{2}}E_2 = \dfrac{9\cdot 10^{9}\cdot 2\cdot 10^{-7}}{0{,}2^{2}}E_2 = 4{,}5\cdot 10^{4}\,\tfrac{\text{В}}{\text{м}}

    On the surface of the ball.

  4. E_3 = \dfrac{kq}{\left(R + S\right)^{2}}E_3 = \dfrac{9\cdot 10^{9}\cdot 2\cdot 10^{-7}}{0{,}3^{2}}E_3 = 2\cdot 10^{4}\,\tfrac{\text{В}}{\text{м}}

AnswerE_1 = 0E_2 = 45\,000\,\tfrac{\text{В}}{\text{м}}E_3 = 20\,000\,\tfrac{\text{В}}{\text{м}}

Position in this set: 5 from 30

Problem #98

Electrostatics: charge and field

There are two kerosene balls with charges of each 4\cdot10^{-7}\,\text{Кл} are at a distance 30\,\text{см}, \varepsilon=2. What is the strength? How to change the distance so that the force increases in 4 times? Given the original distance, find the ratio of the new force to the original force if half the charge of one ball is transferred to another?

Answer options

  1. \left(0{,}008\,\text{Н};\ 0{,}15\,\text{м};\ 0{,}75\,\text{от исходной силы}\right)
  2. \left(0{,}004\,\text{Н};\ 0{,}075\,\text{м};\ 0{,}375\,\text{от исходной силы}\right)
  3. \left(0{,}016\,\text{Н};\ 0{,}3\,\text{м};\ 1{,}5\,\text{от исходной силы}\right)
  4. \left(0{,}08\,\text{Н};\ 1{,}5\,\text{м};\ 7{,}5\,\text{от исходной силы}\right)
Show the walkthroughHide the walkthrough
Given
\left|q_1\right| = \left|q_2\right|
4\cdot 10^{-7}\,\text{Кл}
r
30\,\text{см}0{,}3\,\text{м}
\varepsilon
2
Find

Fr'\dfrac{F''}{F}

  1. F = \dfrac{kq^{2}}{\varepsilon r^{2}}F = \dfrac{9\cdot 10^{9}\cdot\left(4\cdot 10^{-7}\right)^{2}}{2\cdot 0{,}3^{2}}F = 8\cdot 10^{-3}\,\text{Н}
  2. F \sim \dfrac{1}{r^{2}}\ \Rightarrow\ 4F\ \text{при}\ r' = \dfrac{r}{2}r' = \dfrac{0{,}3}{2}r' = 0{,}15\,\text{м}

    The force quadruples if the distance is halved.

  3. q_1' = \dfrac{q}{2},\quad q_2' = \dfrac{3q}{2}\ \Rightarrow\ \dfrac{F''}{F} = \dfrac{q_1'q_2'}{q^{2}} = \dfrac{3}{4}\dfrac{F''}{F} = 0{,}75

    The production of charges has decreased, so the force will be 0.75 of the previous one.

AnswerF = 8\cdot 10^{-3}\,\text{Н}r' = 0{,}15\,\text{м}\dfrac{F''}{F} = 0{,}75

Position in this set: 6 from 30

Problem #99

Electrostatics: potential and field work

In a uniform field of intensity 8000\,\text{Н/Кл} charge 5\cdot10^{-6}\,\text{Кл} moved to 5\,\text{см} along the field. Define the job.

Answer options

  1. 0{,}02\,\text{Дж}
  2. 0{,}002\,\text{Дж}
  3. 0{,}001\,\text{Дж}
  4. 0{,}004\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
E
8000\,\tfrac{\text{Н}}{\text{Кл}}
q
5\cdot 10^{-6}\,\text{Кл}
d
5\,\text{см}5\cdot 10^{-2}\,\text{м}
Find

A

  1. A = Fd = qEdA = 5\cdot 10^{-6}\cdot 8000\cdot 5\cdot 10^{-2}A = 2\cdot 10^{-3}\,\text{Дж} = 2\,\text{мДж}

AnswerA = 2\cdot 10^{-3}\,\text{Дж}

Position in this set: 7 from 30

Problem #100

Electrostatics: charge and field

Distance between plates 0{,}5\,\text{см}, voltage 200\,\text{В}. Between them, in balance, there is a speck of dust that is missing 1000 electrons. Determine the mass of the dust particle.

Answer options

  1. 1{,}28 \cdot 10^{-12}\,\text{кг}
  2. 6{,}4 \cdot 10^{-12}\,\text{кг}
  3. 6{,}4 \cdot 10^{-13}\,\text{кг}
  4. 3{,}2 \cdot 10^{-13}\,\text{кг}
Show the walkthroughHide the walkthrough
Given
d
0{,}5\,\text{см}0{,}5\cdot 10^{-2}\,\text{м}
U
200\,\text{В}
n
1000
e
1{,}6\cdot 10^{-19}\,\text{Кл}
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

m

  1. E = \dfrac{U}{d}E = \dfrac{200}{0{,}5\cdot 10^{-2}}E = 4\cdot 10^{4}\,\tfrac{\text{В}}{\text{м}}
  2. \left|q\right| = ne\left|q\right| = 1000\cdot 1{,}6\cdot 10^{-19}\left|q\right| = 1{,}6\cdot 10^{-16}\,\text{Кл}
  3. qE = mg\ \Rightarrow\ m = \dfrac{qE}{g}m = \dfrac{1{,}6\cdot 10^{-16}\cdot 4\cdot 10^{4}}{10}m = 6{,}4\cdot 10^{-13}\,\text{кг}

    Equilibrium: Electrical force balances gravity.

Answerm = 6{,}4\cdot 10^{-13}\,\text{кг}

Position in this set: 8 from 30

Problem #101

Electrostatics: capacitors

Capacitors 8\,\text{мкФ} and 12\,\text{мкФ} interconnected and connected to 50\,\text{В}. For a series connection, find the total capacitance, the total modulo charge of the plates and the voltage on the first and second capacitors. For a parallel connection, find the total capacitance and charges of the first and second capacitors.

Answer options

  1. \left(2{,}4 \cdot 10^{-6}\,\text{Ф};\ 1{,}2 \cdot 10^{-4}\,\text{Кл};\ 15\,\text{В};\ 10\,\text{В};\ 1 \cdot 10^{-5}\,\text{Ф};\ 2 \cdot 10^{-4}\,\text{Кл};\ 3 \cdot 10^{-4}\,\text{Кл}\right)
  2. \left(9{,}6 \cdot 10^{-6}\,\text{Ф};\ 4{,}8 \cdot 10^{-4}\,\text{Кл};\ 60\,\text{В};\ 40\,\text{В};\ 4 \cdot 10^{-5}\,\text{Ф};\ 8 \cdot 10^{-4}\,\text{Кл};\ 0{,}0012\,\text{Кл}\right)
  3. \left(4{,}8 \cdot 10^{-5}\,\text{Ф};\ 0{,}0024\,\text{Кл};\ 300\,\text{В};\ 200\,\text{В};\ 2 \cdot 10^{-4}\,\text{Ф};\ 0{,}004\,\text{Кл};\ 0{,}006\,\text{Кл}\right)
  4. \left(4{,}8 \cdot 10^{-6}\,\text{Ф};\ 2{,}4 \cdot 10^{-4}\,\text{Кл};\ 30\,\text{В};\ 20\,\text{В};\ 2 \cdot 10^{-5}\,\text{Ф};\ 4 \cdot 10^{-4}\,\text{Кл};\ 6 \cdot 10^{-4}\,\text{Кл}\right)
Show the walkthroughHide the walkthrough
Given
C_1
8\,\text{мкФ}8\cdot 10^{-6}\,\text{Ф}
C_2
12\,\text{мкФ}12\cdot 10^{-6}\,\text{Ф}
U
50\,\text{В}
Find

C_{\text{посл}}qU_1U_2C_{\text{пар}}q_1q_2

  1. C_{\text{посл}} = \dfrac{C_1 C_2}{C_1 + C_2}C_{\text{посл}} = \dfrac{8\cdot 12}{8 + 12}C_{\text{посл}} = 4{,}8\,\text{мкФ} = 4{,}8\cdot 10^{-6}\,\text{Ф}
  2. q = C_{\text{посл}}Uq = 4{,}8\cdot 10^{-6}\cdot 50q = 2{,}4\cdot 10^{-4}\,\text{Кл}

    In a series connection, the charge on both capacitors is the same.

  3. U_1 = \dfrac{q}{C_1},\quad U_2 = \dfrac{q}{C_2}U_1 = \dfrac{2{,}4\cdot 10^{-4}}{8\cdot 10^{-6}};\quad U_2 = \dfrac{2{,}4\cdot 10^{-4}}{12\cdot 10^{-6}}U_1 = 30\,\text{В};\quad U_1 = 20\,\text{В}
  4. C_{\text{пар}} = C_1 + C_2C_{\text{пар}} = 8\cdot 10^{-6} + 12\cdot 10^{-6}C_{\text{пар}} = 2\cdot 10^{-5}\,\text{Ф}
  5. q_1 = C_1 U,\quad q_2 = C_2 Uq_1 = 8\cdot 10^{-6}\cdot 50;\quad q_2 = 12\cdot 10^{-6}\cdot 50q_1 = 4\cdot 10^{-4}\,\text{Кл};\quad q_1 = 6\cdot 10^{-4}\,\text{Кл}

    In a parallel connection, the voltage is the same, not the charge.

AnswerC_{\text{посл}} = 4{,}8\cdot 10^{-6}\,\text{Ф}q = 2{,}4\cdot 10^{-4}\,\text{Кл}U_1 = 30\,\text{В}U_2 = 20\,\text{В}C_{\text{пар}} = 2\cdot 10^{-5}\,\text{Ф}q_1 = 4\cdot 10^{-4}\,\text{Кл}q_2 = 6\cdot 10^{-4}\,\text{Кл}

Position in this set: 9 from 30

Problem #102

Electrostatics: capacitors

Capacitors 100\,\text{пФ} and 400\,\text{пФ} connected to each other. Voltage at first 20\,\text{В}. Determine the voltage on the second and the charges in parallel and series connections.

Answer options

  1. \left(20\,\text{В};\ 2 \cdot 10^{-9}\,\text{Кл};\ 8 \cdot 10^{-9}\,\text{Кл};\ 5\,\text{В};\ 2 \cdot 10^{-9}\,\text{Кл}\right)
  2. \left(10\,\text{В};\ 1 \cdot 10^{-9}\,\text{Кл};\ 4 \cdot 10^{-9}\,\text{Кл};\ 2{,}5\,\text{В};\ 1 \cdot 10^{-9}\,\text{Кл}\right)
  3. \left(40\,\text{В};\ 4 \cdot 10^{-9}\,\text{Кл};\ 1{,}6 \cdot 10^{-8}\,\text{Кл};\ 10\,\text{В};\ 4 \cdot 10^{-9}\,\text{Кл}\right)
  4. \left(200\,\text{В};\ 2 \cdot 10^{-8}\,\text{Кл};\ 8 \cdot 10^{-8}\,\text{Кл};\ 50\,\text{В};\ 2 \cdot 10^{-8}\,\text{Кл}\right)
Show the walkthroughHide the walkthrough
Given
C_1
100\,\text{пФ}100\cdot 10^{-12}\,\text{Ф}
C_2
400\,\text{пФ}400\cdot 10^{-12}\,\text{Ф}
U_1
20\,\text{В}
Find

U_2^{\text{пар}}q_1q_2U_2^{\text{посл}}q^{\text{посл}}

  1. U_2 = U_1 = 20\,\text{В}

    Parallel connection: common voltage.

  2. q_1 = C_1 U_1,\quad q_2 = C_2 U_2q_1 = 100\cdot 10^{-12}\cdot 20;\quad q_2 = 400\cdot 10^{-12}\cdot 20q_1 = 2\cdot 10^{-9}\,\text{Кл};\quad q_1 = 8\cdot 10^{-9}\,\text{Кл}
  3. q = C_1 U_1 = q_2q = 100\cdot 10^{-12}\cdot 20q = 2\cdot 10^{-9}\,\text{Кл}

    Series connection: common charge.

  4. U_2 = \dfrac{q}{C_2}U_2 = \dfrac{2\cdot 10^{-9}}{400\cdot 10^{-12}}U_2^{\text{посл}} = 5\,\text{В}

AnswerU_2^{\text{пар}} = 20\,\text{В}q_1 = 2\cdot 10^{-9}\,\text{Кл}q_2 = 8\cdot 10^{-9}\,\text{Кл}U_2^{\text{посл}} = 5\,\text{В}q^{\text{посл}} = 2\cdot 10^{-9}\,\text{Кл}

Position in this set: 10 from 30

Problem #103

Electrostatics: potential and field work

Distance between plates 5\,\text{см}, voltage 80\,\text{В}. Determine the force acting on the proton and its acceleration.

Answer options

  1. \left(2{,}56 \cdot 10^{-15}\,\text{Н};\ 1{,}53 \cdot 10^{12}\,\text{м/с}^{2}\right)
  2. \left(2{,}56 \cdot 10^{-16}\,\text{Н};\ 1{,}53 \cdot 10^{11}\,\text{м/с}^{2}\right)
  3. \left(1{,}28 \cdot 10^{-16}\,\text{Н};\ 7{,}65 \cdot 10^{10}\,\text{м/с}^{2}\right)
  4. \left(5{,}12 \cdot 10^{-16}\,\text{Н};\ 3{,}06 \cdot 10^{11}\,\text{м/с}^{2}\right)
Show the walkthroughHide the walkthrough
Given
d
5\,\text{см}0{,}05\,\text{м}
U
80\,\text{В}
q
e = 1{,}6\cdot 10^{-19}\,\text{Кл}
m
1{,}67\cdot 10^{-27}\,\text{кг}
Find

Fa

  1. E = \dfrac{U}{d}E = \dfrac{80}{0{,}05}E = 1600\,\tfrac{\text{В}}{\text{м}}
  2. F = qEF = 1{,}6\cdot 10^{-19}\cdot 1600F = 2{,}56\cdot 10^{-16}\,\text{Н}
  3. a = \dfrac{F}{m}a = \dfrac{2{,}56\cdot 10^{-16}}{1{,}67\cdot 10^{-27}}a = 1{,}53\cdot 10^{11}\,\tfrac{\text{м}}{\text{с}^{2}}

AnswerF = 2{,}56\cdot 10^{-16}\,\text{Н}a = 1{,}53\cdot 10^{11}\,\tfrac{\text{м}}{\text{с}^{2}}

Position in this set: 11 from 30

Problem #104

Electrostatics: capacitors

Capacitors 300\,\text{пФ} and 700\,\text{пФ} connected in series and connected to 200\,\text{В}. Determine the voltage across each capacitor.

Answer options

  1. \left(280\,\text{В};\ 120\,\text{В}\right)
  2. \left(1400\,\text{В};\ 600\,\text{В}\right)
  3. \left(140\,\text{В};\ 60\,\text{В}\right)
  4. \left(70\,\text{В};\ 30\,\text{В}\right)
Show the walkthroughHide the walkthrough
Given
C_1
300\,\text{пФ}
C_2
700\,\text{пФ}
U
200\,\text{В}
Find

U_1U_2

  1. C = \dfrac{C_1 C_2}{C_1 + C_2}C = \dfrac{300\cdot 700}{300 + 700}C = 210\,\text{пФ}
  2. q = CUq = 210\cdot 10^{-12}\cdot 200q = 4{,}2\cdot 10^{-8}\,\text{Кл}
  3. U_1 = \dfrac{q}{C_1}U_1 = \dfrac{4{,}2\cdot 10^{-8}}{300\cdot 10^{-12}}U_1 = 140\,\text{В}
  4. U_2 = U - U_1U_2 = 200 - 140U_2 = 60\,\text{В}

    Check: q/C_2 gives the same value.

AnswerU_1 = 140\,\text{В}U_2 = 60\,\text{В}

Position in this set: 12 from 30

Problem #105

Electrostatics: capacitors

Capacitors 8\,\text{мкФ} and 2\,\text{мкФ} connected to 150\,\text{В}. For a parallel connection, find the total capacitance and charges of the first and second capacitors. For a series connection, find the total capacitance, the total module charge of the plates and the voltage on the second capacitor.

Answer options

  1. \left(5 \cdot 10^{-6}\,\text{Ф};\ 6 \cdot 10^{-4}\,\text{Кл};\ 1{,}5 \cdot 10^{-4}\,\text{Кл};\ 8 \cdot 10^{-7}\,\text{Ф};\ 1{,}2 \cdot 10^{-4}\,\text{Кл};\ 60\,\text{В}\right)
  2. \left(2 \cdot 10^{-5}\,\text{Ф};\ 0{,}0024\,\text{Кл};\ 6 \cdot 10^{-4}\,\text{Кл};\ 3{,}2 \cdot 10^{-6}\,\text{Ф};\ 4{,}8 \cdot 10^{-4}\,\text{Кл};\ 240\,\text{В}\right)
  3. \left(1 \cdot 10^{-4}\,\text{Ф};\ 0{,}012\,\text{Кл};\ 0{,}003\,\text{Кл};\ 1{,}6 \cdot 10^{-5}\,\text{Ф};\ 0{,}0024\,\text{Кл};\ 1200\,\text{В}\right)
  4. \left(1 \cdot 10^{-5}\,\text{Ф};\ 0{,}0012\,\text{Кл};\ 3 \cdot 10^{-4}\,\text{Кл};\ 1{,}6 \cdot 10^{-6}\,\text{Ф};\ 2{,}4 \cdot 10^{-4}\,\text{Кл};\ 120\,\text{В}\right)
Show the walkthroughHide the walkthrough
Given
C_1
8\,\text{мкФ}8\cdot 10^{-6}\,\text{Ф}
C_2
2\,\text{мкФ}2\cdot 10^{-6}\,\text{Ф}
U
150\,\text{В}
Find

C_{\text{пар}}q_1q_2C_{\text{посл}}qU_2

  1. C_{\text{пар}} = C_1 + C_2C_{\text{пар}} = 8\cdot 10^{-6} + 2\cdot 10^{-6}C_{\text{пар}} = 10^{-5}\,\text{Ф}
  2. q_1 = C_1 U,\quad q_2 = C_2 Uq_1 = 8\cdot 10^{-6}\cdot 150;\quad q_2 = 2\cdot 10^{-6}\cdot 150q_1 = 1{,}2\cdot 10^{-3}\,\text{Кл};\quad q_1 = 3\cdot 10^{-4}\,\text{Кл}
  3. C_{\text{посл}} = \dfrac{C_1 C_2}{C_1 + C_2}C_{\text{посл}} = \dfrac{8\cdot10^{-6}\cdot2\cdot10^{-6}}{8\cdot10^{-6}+2\cdot10^{-6}}C_{\text{посл}} = 1{,}6\cdot 10^{-6}\,\text{Ф}
  4. q = C_{\text{посл}}Uq = 1{,}6\cdot 10^{-6}\cdot 150q = 2{,}4\cdot 10^{-4}\,\text{Кл}
  5. U_1 = \dfrac{q}{C_1},\quad U_2 = \dfrac{q}{C_2}\dfrac{2{,}4\cdot 10^{-4}}{8\cdot 10^{-6}} = 30\,\text{В};\quad U_2 = \dfrac{2{,}4\cdot 10^{-4}}{2\cdot 10^{-6}}U_2 = 120\,\text{В}

    The sum of the voltages gives 150 V - this is a check of the calculation.

AnswerC_{\text{пар}} = 10^{-5}\,\text{Ф}q_1 = 1{,}2\cdot 10^{-3}\,\text{Кл}q_2 = 3\cdot 10^{-4}\,\text{Кл}C_{\text{посл}} = 1{,}6\cdot 10^{-6}\,\text{Ф}q = 2{,}4\cdot 10^{-4}\,\text{Кл}U_2 = 120\,\text{В}

Position in this set: 13 from 30

Problem #207

Charge and electric field

Per charge 0{,}000002 Cl acts on a uniform field strength 1000 N/Cl. Find the magnitude of the force using the formula F = |q|E.

Answer options

  1. 0{,}002\,\text{Н}
  2. 0{,}004\,\text{Н}
  3. 0{,}001\,\text{Н}
  4. 0{,}006\,\text{Н}
Show the walkthroughHide the walkthrough
Given
q
0{,}000002\,\text{Кл}
E
1000\,\text{Н/Кл}
Find

F

  1. F=qEF=0{,}000002 \cdot 1000F=0{,}002\,\text{Н}

    We take the module of positive charge from the condition.

Answer0{,}002\,\text{Н}

Position in this set: 14 from 30

Problem #208

Charge and electric field

Per charge 0{,}000003 Cl acts on a uniform field strength 2000 N/Cl. Find the magnitude of the force using the formula F = |q|E.

Answer options

  1. 0{,}012\,\text{Н}
  2. 0{,}006\,\text{Н}
  3. 0{,}003\,\text{Н}
  4. 0{,}018\,\text{Н}
Show the walkthroughHide the walkthrough
Given
q
0{,}000003\,\text{Кл}
E
2000\,\text{Н/Кл}
Find

F

  1. F=qEF=0{,}000003 \cdot 2000F=0{,}006\,\text{Н}

    We take the module of positive charge from the condition.

Answer0{,}006\,\text{Н}

Position in this set: 15 from 30

Problem #209

Charge and electric field

On a positive charge 2 µC force acts 0{,}006 N. Find the field strength in N/C.

Answer options

  1. 6000\,\text{Н/Кл}
  2. 3000\,\text{Н/Кл}
  3. 1500\,\text{Н/Кл}
  4. 9000\,\text{Н/Кл}
Show the walkthroughHide the walkthrough
Given
q
2\,\text{мкКл}
F
0{,}006\,\text{Н}
Find

E

  1. Q=q\cdot10^{-6}Q=\frac{2}{1000000}Q=0{,}000002\,\text{Кл}

    A microcoulomb is a millionth of a coulomb.

  2. E=\frac{F}{Q}E=\frac{0{,}006}{0{,}000002}E=3000\,\text{Н/Кл}

    We divide the force by the test charge.

Answer3000\,\text{Н/Кл}

Position in this set: 16 from 30

Problem #210

Charge and electric field

On a positive charge 5 µC force acts 0{,}02 N. Find the field strength in N/C.

Answer options

  1. 8000\,\text{Н/Кл}
  2. 2000\,\text{Н/Кл}
  3. 4000\,\text{Н/Кл}
  4. 12000\,\text{Н/Кл}
Show the walkthroughHide the walkthrough
Given
q
5\,\text{мкКл}
F
0{,}02\,\text{Н}
Find

E

  1. Q=q\cdot10^{-6}Q=\frac{5}{1000000}Q=0{,}000005\,\text{Кл}

    A microcoulomb is a millionth of a coulomb.

  2. E=\frac{F}{Q}E=\frac{0{,}02}{0{,}000005}E=4000\,\text{Н/Кл}

    We divide the force by the test charge.

Answer4000\,\text{Н/Кл}

Position in this set: 17 from 30

Problem #211

Charge and electric field

Initially, two point charges interacted with the force 0{,}8 N. The first charge was increased by 2 once, twice 3 times, and the distance increased by 2 times. Find a new strength. Use the relationship F ∝ q₁q₂/r².

Answer options

  1. 2{,}4\,\text{Н}
  2. 0{,}6\,\text{Н}
  3. 1{,}2\,\text{Н}
  4. 3{,}6\,\text{Н}
Show the walkthroughHide the walkthrough
Given
F
0{,}8\,\text{Н}
a
2\,\text{раз}
b
3\,\text{раз}
r
2\,\text{раз}
Find

F_{\text{нов}}

  1. k=abk=2 \cdot 3k=6\,\text{раз}

    The product of charges increases by ab times.

  2. d=r^2d=2 \cdot 2d=4\,\text{раз}

    Increasing the distance weakens the force in proportion to the square.

  3. F_{\text{нов}}=F\frac{k}{d}F_{\text{нов}}=\frac{0{,}8 \cdot 6}{4}F_{\text{нов}}=1{,}2\,\text{Н}

    Let's combine both changes.

Answer1{,}2\,\text{Н}

Position in this set: 18 from 30

Problem #212

Charge and electric field

Initially, two point charges interacted with the force 1{,}2 N. The first charge was increased by 3 once, twice 2 times, and the distance increased by 3 times. Find a new strength. Use the relationship F ∝ q₁q₂/r².

Answer options

  1. 1{,}6\,\text{Н}
  2. 0{,}4\,\text{Н}
  3. 2{,}4\,\text{Н}
  4. 0{,}8\,\text{Н}
Show the walkthroughHide the walkthrough
Given
F
1{,}2\,\text{Н}
a
3\,\text{раз}
b
2\,\text{раз}
r
3\,\text{раз}
Find

F_{\text{нов}}

  1. k=abk=3 \cdot 2k=6\,\text{раз}

    The product of charges increases by ab times.

  2. d=r^2d=3 \cdot 3d=9\,\text{раз}

    Increasing the distance weakens the force in proportion to the square.

  3. F_{\text{нов}}=F\frac{k}{d}F_{\text{нов}}=\frac{1{,}2 \cdot 6}{9}F_{\text{нов}}=0{,}8\,\text{Н}

    Let's combine both changes.

Answer0{,}8\,\text{Н}

Position in this set: 19 from 30

Problem #213

Potential and voltage

Positive charge 0{,}002 Cl passes through potential difference 100 In the direction of the field strength. Find the work of the field.

Answer options

  1. 0{,}4\,\text{Дж}
  2. 0{,}2\,\text{Дж}
  3. 0{,}1\,\text{Дж}
  4. 0{,}6\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
q
0{,}002\,\text{Кл}
U
100\,\text{В}
Find

A

  1. A=qUA=0{,}002 \cdot 100A=0{,}2\,\text{Дж}

    The potential decreases, the field work is positive.

Answer0{,}2\,\text{Дж}

Position in this set: 20 from 30

Problem #214

Potential and voltage

Positive charge 0{,}003 Cl passes through potential difference 200 In the direction of the field strength. Find the work of the field.

Answer options

  1. 1{,}2\,\text{Дж}
  2. 0{,}3\,\text{Дж}
  3. 0{,}6\,\text{Дж}
  4. 1{,}8\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
q
0{,}003\,\text{Кл}
U
200\,\text{В}
Find

A

  1. A=qUA=0{,}003 \cdot 200A=0{,}6\,\text{Дж}

    The potential decreases, the field work is positive.

Answer0{,}6\,\text{Дж}

Position in this set: 21 from 30

Problem #215

Potential and voltage

The field does work 4 mJ over charge 20 µC Find the potential difference passed through.

Answer options

  1. 400\,\text{В}
  2. 100\,\text{В}
  3. 200\,\text{В}
  4. 600\,\text{В}
Show the walkthroughHide the walkthrough
Given
A
4\,\text{мДж}
q
20\,\text{мкКл}
Find

U

  1. a=A\cdot10^{-3}a=\frac{4}{1000}a=0{,}004\,\text{Дж}

    Convert millijoules to joules.

  2. Q=q\cdot10^{-6}Q=\frac{20}{1000000}Q=0{,}00002\,\text{Кл}

    Convert microcoulombs to coulombs.

  3. U=\frac{a}{Q}U=\frac{0{,}004}{0{,}00002}U=200\,\text{В}

    We divide the work into the charge.

Answer200\,\text{В}

Position in this set: 22 from 30

Problem #216

Potential and voltage

The field does work 6 mJ over charge 30 µC Find the potential difference passed through.

Answer options

  1. 400\,\text{В}
  2. 100\,\text{В}
  3. 600\,\text{В}
  4. 200\,\text{В}
Show the walkthroughHide the walkthrough
Given
A
6\,\text{мДж}
q
30\,\text{мкКл}
Find

U

  1. a=A\cdot10^{-3}a=\frac{6}{1000}a=0{,}006\,\text{Дж}

    Convert millijoules to joules.

  2. Q=q\cdot10^{-6}Q=\frac{30}{1000000}Q=0{,}00003\,\text{Кл}

    Convert microcoulombs to coulombs.

  3. U=\frac{a}{Q}U=\frac{0{,}006}{0{,}00003}U=200\,\text{В}

    We divide the work into the charge.

Answer200\,\text{В}

Position in this set: 23 from 30

Problem #217

Potential and voltage

Particle with mass 2\cdot 10^{-9} kg with positive charge 0{,}000001 Cl is accelerated from rest by a potential difference 100 B. Find the speed. There are no other forces; the movement is non-relativistic.

Answer options

  1. 632{,}456\,\text{м/с}
  2. 158{,}114\,\text{м/с}
  3. 948{,}683\,\text{м/с}
  4. 316{,}228\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
m
2\cdot 10^{-9}\,\text{кг}
q
0{,}000001\,\text{Кл}
U
100\,\text{В}
Find

v

  1. A=qUA=0{,}000001 \cdot 100A=0{,}0001\,\text{Дж}

    Field work increases kinetic energy.

  2. v^2=\frac{2A}{m}v^2=\frac{2 \cdot 0{,}0001}{2\cdot 10^{-9}}v^2=100000\,\text{м}^{2}\text{/с}^{2}

    We use mv²/2 = A.

  3. v=\sqrt{\frac{2qU}{m}}v=\sqrt{100000}v=316{,}228\,\text{м/с}

    Finding the final velocity modulus.

Answer316{,}228\,\text{м/с}

Position in this set: 24 from 30

Problem #218

Potential and voltage

Particle with mass 4\cdot 10^{-9} kg with positive charge 0{,}000002 Cl is accelerated from rest by a potential difference 200 B. Find the speed. There are no other forces; the movement is non-relativistic.

Answer options

  1. 447{,}214\,\text{м/с}
  2. 894{,}427\,\text{м/с}
  3. 223{,}607\,\text{м/с}
  4. 1341{,}64\,\text{м/с}
Show the walkthroughHide the walkthrough
Given
m
4\cdot 10^{-9}\,\text{кг}
q
0{,}000002\,\text{Кл}
U
200\,\text{В}
Find

v

  1. A=qUA=0{,}000002 \cdot 200A=0{,}0004\,\text{Дж}

    Field work increases kinetic energy.

  2. v^2=\frac{2A}{m}v^2=\frac{2 \cdot 0{,}0004}{4\cdot 10^{-9}}v^2=200000\,\text{м}^{2}\text{/с}^{2}

    We use mv²/2 = A.

  3. v=\sqrt{\frac{2qU}{m}}v=\sqrt{200000}v=447{,}214\,\text{м/с}

    Finding the final velocity modulus.

Answer447{,}214\,\text{м/с}

Position in this set: 25 from 30

Problem #219

Capacitors

Capacitor capacity 4 µF, voltage 12 B. Find the charge in microcoulombs.

Answer options

  1. 96\,\text{мкКл}
  2. 24\,\text{мкКл}
  3. 48\,\text{мкКл}
  4. 144\,\text{мкКл}
Show the walkthroughHide the walkthrough
Given
C
4\,\text{мкФ}
U
12\,\text{В}
Find

q

  1. q=CUq=4 \cdot 12q=48\,\text{мкКл}

    Microfarads when multiplied by volts give microcoulombs.

Answer48\,\text{мкКл}

Position in this set: 26 from 30

Problem #220

Capacitors

Capacitor capacity 6 µF, voltage 20 B. Find the charge in microcoulombs.

Answer options

  1. 240\,\text{мкКл}
  2. 60\,\text{мкКл}
  3. 360\,\text{мкКл}
  4. 120\,\text{мкКл}
Show the walkthroughHide the walkthrough
Given
C
6\,\text{мкФ}
U
20\,\text{В}
Find

q

  1. q=CUq=6 \cdot 20q=120\,\text{мкКл}

    Microfarads when multiplied by volts give microcoulombs.

Answer120\,\text{мкКл}

Position in this set: 27 from 30

Problem #221

Capacitors

Capacitor with capacity 100 uF charged to 20 B. Find the stored energy in joules.

Answer options

  1. 0{,}04\,\text{Дж}
  2. 0{,}01\,\text{Дж}
  3. 0{,}06\,\text{Дж}
  4. 0{,}02\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
C
100\,\text{мкФ}
U
20\,\text{В}
Find

W

  1. c=C\cdot10^{-6}c=\frac{100}{1000000}c=0{,}0001\,\text{Ф}

    Convert the capacitance to farads.

  2. W=\frac{cU^2}{2}W=\frac{0{,}0001 \cdot {20}^{2}}{2}W=0{,}02\,\text{Дж}

    Energy depends on the square of voltage.

Answer0{,}02\,\text{Дж}

Position in this set: 28 from 30

Problem #222

Capacitors

Capacitor with capacity 200 uF charged to 30 B. Find the stored energy in joules.

Answer options

  1. 0{,}09\,\text{Дж}
  2. 0{,}18\,\text{Дж}
  3. 0{,}045\,\text{Дж}
  4. 0{,}27\,\text{Дж}
Show the walkthroughHide the walkthrough
Given
C
200\,\text{мкФ}
U
30\,\text{В}
Find

W

  1. c=C\cdot10^{-6}c=\frac{200}{1000000}c=0{,}0002\,\text{Ф}

    Convert the capacitance to farads.

  2. W=\frac{cU^2}{2}W=\frac{0{,}0002 \cdot {30}^{2}}{2}W=0{,}09\,\text{Дж}

    Energy depends on the square of voltage.

Answer0{,}09\,\text{Дж}

Position in this set: 29 from 30

Problem #223

Capacitors

Capacitor 4 uF charged to 30 V, capacitor 8 μF is not charged. They were connected in parallel, turning off the source. Find the final voltage.

Answer options

  1. 10\,\text{В}
  2. 20\,\text{В}
  3. 5\,\text{В}
  4. 30\,\text{В}
Show the walkthroughHide the walkthrough
Given
C_1
4\,\text{мкФ}
C_2
8\,\text{мкФ}
U
30\,\text{В}
Find

V

  1. q=C_1Uq=4 \cdot 30q=120\,\text{мкКл}

    The total charge of the corresponding plates is maintained.

  2. C=C_1+C_2C=4 + 8C=12\,\text{мкФ}

    Parallel containers are stacked.

  3. V=\frac{q}{C}V=\frac{120}{12}V=10\,\text{В}

    When combined, energy is partially converted into other forms.

Answer10\,\text{В}

Position in this set: 30 from 30

Problem #224

Capacitors

Capacitor 6 uF charged to 18 V, capacitor 3 μF is not charged. They were connected in parallel, turning off the source. Find the final voltage.

Answer options

  1. 24\,\text{В}
  2. 12\,\text{В}
  3. 6\,\text{В}
  4. 36\,\text{В}
Show the walkthroughHide the walkthrough
Given
C_1
6\,\text{мкФ}
C_2
3\,\text{мкФ}
U
18\,\text{В}
Find

V

  1. q=C_1Uq=6 \cdot 18q=108\,\text{мкКл}

    The total charge of the corresponding plates is maintained.

  2. C=C_1+C_2C=6 + 3C=9\,\text{мкФ}

    Parallel containers are stacked.

  3. V=\frac{q}{C}V=\frac{108}{9}V=12\,\text{В}

    When combined, energy is partially converted into other forms.

Answer12\,\text{В}

The theory behind these problems

Work through the topic if a problem will not give in.

Charge and electric fieldCharge conservation, Coulomb's law, electric field strength, and superposition.Potential and voltageWork done by the field, potential energy, electric potential, and potential difference.CapacitorsCapacitance, parallel-plate capacitors, energy, and connections.

Formulas by topic

Look up symbols and units of measurement in the reference.

Electric chargeq, Q · КлElectrical voltageU · ВElectric field strengthE · В/м, Н/КлElectric potentialφ · ВWorkA · ДжDistance and pathS, l · мCapacitanceC · ФEnergyE, W · Дж