Electrostatics Worksheet
Topics:
Redistribution of charges in spheres, Van de Graaff Generator, Action of Points, Earthing, Charge Flow
Section A: Theory Questions (10)
1. Explain redistribution of charges when two conducting spheres are connected.
2. Why does charge always reside on the outer surface of a conductor?
3. What is meant by action of points?
4. Explain why sharp points discharge charges easily.
5. State the principle of Van de Graaff generator.
6. Why does charge move from inner sphere to outer sphere in a hollow conductor?
7. What happens to potential when a charged sphere is earthed?
8. Define earthing and its purpose.
9. Why is electric field inside a conductor zero?
10. What happens to charge distribution when a conductor is placed in an external electric field?
Section B: Numerical Questions (10)
11. Two spheres of radii \(R_1 = 2 cm\), \(R_2 = 4 cm\) carry charges \(2\mu C\) and \(4\mu C\). Find final charges after connection.
12. A sphere of radius \(R\) has charge \(Q\). Find its potential.
13. Two identical spheres, one charged \(6\mu C\), other neutral. After contact, find final charge on each.
14. A sphere is earthed. Initial charge is \(10\mu C\). Find final charge.
15. Potential of a sphere is \(V = 1000V\), radius \(0.1m\). Find charge.
16. A Van de Graaff generator sphere has radius \(0.5m\) and potential \(10^6 V\). Find charge.
17. Charge density at sharp point is higher than flat surface. Explain using formula.
18. Two spheres connected by wire: radii ratio \(1:2\). Total charge \(9\mu C\). Find charges.
19. A sphere is placed in electric field \(E\). What is field inside?
20. A hollow conductor has inner radius \(a\), outer radius \(b\). Charge \(Q\). Find electric field inside cavity.
Answer Key
1. Charges redistribute to equalize potential
or
Charges flow from higher voltage to lower voltage until voltages are equal
2. Due to repulsion of charges and reach an electrostatic equilibrium
3. Charge concentration at sharp points, higher surface charge density leads to leakage of charges called Action of Points
4. High electric field causes ionization
5. Charge accumulation using moving belt
6. Inner sphere is always at a higher potential than the outer sphere, so charge always flows from inner sphere to outer sphere
7. Potential becomes zero
8. Connecting to earth to neutralize charge
9. Charged conductor: Electrostatic equilibrium, forces cancel each other
Uncharged conductor in External electric field: E_ext. = E_int., E_int. is equal and opposite to E_ext.
E_net=E_ext.-E_int.=0
10. Induced charges appear, Uncharged conductor in External electric field: E_ext. = E_int., E_int. is equal and opposite to E_ext.
E_net=E_ext.-E_int.=0
11. \(q_1'=2\mu C, q_2'=4\mu C\)
12. \(V = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}\)
13. \(3\mu C\) each
14. Zero
15. \(11.1 \,nC\)
16. \(55.5 \, \mu C\)
17. \(\sigma = \dfrac{Q}{A}\)
18. \(q_1'=3\,\mu C, q_2'=6\,\mu C\)
19. Zero
20. Zero
or
Charges flow from higher voltage to lower voltage until voltages are equal
2. Due to repulsion of charges and reach an electrostatic equilibrium
3. Charge concentration at sharp points, higher surface charge density leads to leakage of charges called Action of Points
4. High electric field causes ionization
5. Charge accumulation using moving belt
6. Inner sphere is always at a higher potential than the outer sphere, so charge always flows from inner sphere to outer sphere
7. Potential becomes zero
8. Connecting to earth to neutralize charge
9. Charged conductor: Electrostatic equilibrium, forces cancel each other
Uncharged conductor in External electric field: E_ext. = E_int., E_int. is equal and opposite to E_ext.
E_net=E_ext.-E_int.=0
10. Induced charges appear, Uncharged conductor in External electric field: E_ext. = E_int., E_int. is equal and opposite to E_ext.
E_net=E_ext.-E_int.=0
11. \(q_1'=2\mu C, q_2'=4\mu C\)
12. \(V = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R}\)
13. \(3\mu C\) each
14. Zero
15. \(11.1 \,nC\)
16. \(55.5 \, \mu C\)
17. \(\sigma = \dfrac{Q}{A}\)
18. \(q_1'=3\,\mu C, q_2'=6\,\mu C\)
19. Zero
20. Zero
Detailed Solutions
Q11:
Method 1:
\[ V_1=k\frac{q_1}{r_1} = k\frac{2\mu C}{2 cm}\\ \\
V_2=k\frac{q_2}{r_2} = k\frac{4\mu C}{4 cm}\\ \] Since Voltages are equal, there will not be any charge flow.
final charges = \(2\mu C, 4\mu C\)
Method 2: After connecting spheres, charges flow from higher voltage to lower voltage, until equilibrium voltage V is reached.
\[ V=k\frac{q_1'}{r_1}=k\frac{q_2'}{r_2}\\ \frac{q_1'}{r_1}=\frac{q_2'}{r_2}\\ \frac{q_1'}{r_1}=const.\\ q'\propto r\\ \sigma \propto \frac{1}{r}\\ \] Important results,
At equilibrium charge is proportional to radius,
At equilibrium surface charge density is inversely proportional to radius.
\[ q_1'\propto 2 cm \\ q_2'\propto 4 cm \\ q_2'=2 \times q_1'\\ \] By Charge conservation: \[ q_{before}=q_{after}\\ q_1+q_2=q_1'+q_2'\\ 2\mu C + 4\mu C =q_1' + 2 \times q_1'\\ q_1'= 2 \mu C\\ q_2'= 4 \mu C\\ \] Charges are unchanged as seen earlier because there was no voltage / potential difference to begin with.
Method 1:
\[ V_1=k\frac{q_1}{r_1} = k\frac{2\mu C}{2 cm}\\ \\
V_2=k\frac{q_2}{r_2} = k\frac{4\mu C}{4 cm}\\ \] Since Voltages are equal, there will not be any charge flow.
final charges = \(2\mu C, 4\mu C\)
Method 2: After connecting spheres, charges flow from higher voltage to lower voltage, until equilibrium voltage V is reached.
\[ V=k\frac{q_1'}{r_1}=k\frac{q_2'}{r_2}\\ \frac{q_1'}{r_1}=\frac{q_2'}{r_2}\\ \frac{q_1'}{r_1}=const.\\ q'\propto r\\ \sigma \propto \frac{1}{r}\\ \] Important results,
At equilibrium charge is proportional to radius,
At equilibrium surface charge density is inversely proportional to radius.
\[ q_1'\propto 2 cm \\ q_2'\propto 4 cm \\ q_2'=2 \times q_1'\\ \] By Charge conservation: \[ q_{before}=q_{after}\\ q_1+q_2=q_1'+q_2'\\ 2\mu C + 4\mu C =q_1' + 2 \times q_1'\\ q_1'= 2 \mu C\\ q_2'= 4 \mu C\\ \] Charges are unchanged as seen earlier because there was no voltage / potential difference to begin with.
Q12:
\[ V = \frac{1}{4\pi\epsilon_0}\frac{Q}{R} \]
\[ V = \frac{1}{4\pi\epsilon_0}\frac{Q}{R} \]
Q13:
Equal sharing → each gets: \[ \frac{6}{2} = 3\mu C \]
Equal sharing → each gets: \[ \frac{6}{2} = 3\mu C \]
Q14:
Earthing → charge flows to earth → final = 0
Earthing → charge flows to earth → final = 0
Q15:
\[ Q = 4\pi\epsilon_0 R V=\dfrac{1}{9\times10^9}\times0.1\times10^3=1.11\times10^{-8}=11.1\times10^{-9}=11.1 \,nC \]
\[ Q = 4\pi\epsilon_0 R V=\dfrac{1}{9\times10^9}\times0.1\times10^3=1.11\times10^{-8}=11.1\times10^{-9}=11.1 \,nC \]
Q16:
\[ Q = 4\pi\epsilon_0 R V=\dfrac{1}{9\times10^9}\times0.5\times10^6=\dfrac{5}{9}\times10^{-4}=0.55\times10^{-4}=55.5 \,\mu C \]
\[ Q = 4\pi\epsilon_0 R V=\dfrac{1}{9\times10^9}\times0.5\times10^6=\dfrac{5}{9}\times10^{-4}=0.55\times10^{-4}=55.5 \,\mu C \]
Q17:
\[ \sigma = \frac{Q}{A} \] Sharp points → small area → large \(\sigma\)
\[ \sigma = \frac{Q}{A} \] Sharp points → small area → large \(\sigma\)
Q18:
Refer to detailed explanation in Q11. \[ q'\propto r\\ q_1'\propto r_1\,\,\,\,q_1'\propto 1\\ q_2'\propto r_2\,\,\,\,q_2'\propto 2\\ .\\ q_2'=2\times q_1'\\ q_1'+q_2'=9\,\mu C\\ \] Solve to get, \[ q_1'=3 \,\mu C \\ q_2'=6 \,\mu C \\ \]
Refer to detailed explanation in Q11. \[ q'\propto r\\ q_1'\propto r_1\,\,\,\,q_1'\propto 1\\ q_2'\propto r_2\,\,\,\,q_2'\propto 2\\ .\\ q_2'=2\times q_1'\\ q_1'+q_2'=9\,\mu C\\ \] Solve to get, \[ q_1'=3 \,\mu C \\ q_2'=6 \,\mu C \\ \]
Q19:
Inside conductor: \[ E = 0 \]
Inside conductor: \[ E = 0 \]
Q20:
Inside cavity: \[ E = 0 \]
Inside cavity: \[ E = 0 \]
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