Numerical Problems of chapter 13: Electrostatics

Statements:

1. How many electrons are removed from a metallic sphere to get a charge of 100 pC?

2. Two point charges of 1.2 pC and 2.5 pC are separated by a 50 cm distance. Find the magnitude of the force between the two point charges.

3. The force of repulsion between two identical point charges is 100 N when the distance between them is 40 cm. What will be the force of repulsion between the two charges if the distance is doubled?

4. Two identical charges, separated by 10 cm, experience an electrostatic force of 250 N. Find the magnitude of the charges.

5. 90 J of work is done to move a 2 C charge between two points having a potential difference. The point at lower potential has an electric potential of 13 V. Find the electric potential of the other point.

6. Two capacitors of capacitances 5 μF and 10 μF are connected in parallel with a 20 V battery. Find:
a. Equivalent capacitance of the combination
b. Total charge stored on the combination
c. Charge on each capacitor
d. Potential difference across each capacitor

7. Two capacitors of capacitances 3 μF and 6 μF are connected in series with a 10 V battery. Find:
a. Equivalent capacitance of the combination
b. Total charge stored on the combination
c. Charge on each capacitor
d. Potential difference across each capacitor

8. If a capacitor stores 10 C charge when connected with a 200 V battery, how much charge will be stored on it if connected with a 45 V battery?

1. How many electrons are removed from a metallic sphere to get a charge of 100 pC?

The charge q on an object is related to the number of electrons removed (or added) using the formula:

q=n⋅e

where:

• $q$ is the total charge,
• $n$ is the number of electrons removed,
• $e$ is the charge of one electron, e=1.6×10⁻¹⁹ C

We are given:

q=100 pC

=100×10⁻¹² C

We can solve for n (the number of electrons):

n=q/e

=100×10⁻¹²/1.6×10⁻¹⁹

=6.25×10⁶

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2. Two point charges of 1.2 pC and 2.5 pC are separated by a 50 cm distance. Find the magnitude of the force between the two point charges.

The force between two point charges is given by Coulomb’s law:

F=ke⋅∣q1q2∣/r2​

Substitute the given values:

• ke=9×10⁹ N⋅m²/C²
• q1=1.2×10⁻¹² C
• q2=2.5×10⁻¹² C
• r=0.5 m

F=ke​⋅∣q1​q2​∣/r²​

Now, calculate the force:

F=9×10⁹⋅∣1.2×10⁻¹²⋅2.5×10⁻¹²∣/(0.5)²=1.08×10⁻³ N

So, the force between the two point charges is 1.08 mN.

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3. The force of repulsion between two identical point charges is 100 N when the distance between them is 40 cm. What will be the force of repulsion between the two charges if the distance is doubled?

Since Coulomb’s law shows that force is inversely proportional to the square of the distance,

F∝1/r2​

if the distance is doubled, the force will reduce by a factor of 2²=4

Given the initial force of 100 N:

The new force F2​ will be:

F2=F1/4​​

F2=100/4=25 N

So, the new force will be 25 N when the distance is doubled.

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4. Two identical charges, separated by 10 cm, experience an electrostatic force of 250 N. Find the magnitude of the charges.

Using Coulomb’s law:

F=ke⋅q2/r2​

Rearranging to solve for q:

q=√F⋅r2/ke​​

Substitute the known values:

• F=250 N
• r=0.1 m
• ke=9×10⁹N⋅m²/C²

Now, calculate $q$:

q=√250⋅(0.1)²/9×10⁹

=5.28×10⁻⁴ C

So, the magnitude of each charge is approximately 0.528 mC.

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5. 90 J of work is done to move a 2 C charge between two points having a potential difference. The point at lower potential has an electric potential of 13 V. Find the electric potential of the other point.

The work done to move a charge across a potential difference is:

W=q⋅ΔV

where ΔV=V_final−V_initial

We can solve for V_final

ΔV=W/q,  V_final=V_initial+ΔV

Substitute the known values:

• W=90 J,
• q=2 C,
• V_initial=13 V

Now, calculate V_final:

ΔV=90/2=45 V    ,V_final=13+45=58 V

So, the electric potential at the other point is 58 V.

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6. Two capacitors of capacitances 5 μF and 10 μF are connected in parallel with a 20 V battery.

a. Equivalent capacitance of the combination

For capacitors in parallel:

Ceq=C1+C2=5 μF+10 μF=15 μF

b. Total charge stored on the combination

The total charge is:

Q_total=C_eq⋅V=15×10−6 F×20 V=300×10−6 C=300 μC

c. Charge on each capacitor

For the 5 μF capacitor:

Q1=C1⋅V=5×10⁻⁶×20=100 μC

For the 10 μF capacitor:

Q2=C2⋅V=10×10⁻⁶×20=200 μC

d. Potential difference across each capacitor

Since they are in parallel, the potential difference across each capacitor is the same as the battery voltage, i.e., 20 V.

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7. Two capacitors of capacitances 3 μF and 6 μF are connected in series with a 10 V battery.

a. Equivalent capacitance of the combination

For capacitors in series:

1/Ceq=1/C1+1/C2

1/Ceq=1/3 μF+1/6 μF=1/2 μF⇒Ceq=2 μF

b. Total charge stored on the combination

The total charge is:

Q_total=C_eq⋅V =2×10⁻⁶×10=20 μC

c. Charge on each capacitor

The charge on each capacitor is the same in a series combination, so the charge on both capacitors is 20 μC.

d. Potential difference across each capacitor

For the 3 μF capacitor:

V1=Q/C1=20×10⁻⁶/3×10⁻⁶=6.67 V

For the 6 μF capacitor:

V2=Q/C2=20×10⁻⁶/6×10⁻⁶=3.33 V

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8. If a capacitor stores 10 C charge when connected with a 200 V battery, how much charge will be stored on it if connected with a 45 V battery?

The capacitance of the capacitor is:

C=Q/V

=10 C/200 V

$$\text{F}$$

Now, with the 45 V battery, the charge stored will be:

Q=C⋅V=0.05×45=2.25 C

So, the charge stored with the 45 V battery is 2.25 C.