NEET Electrostatic Potential and Capacitance — Set 10

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Question 1

A charged particle is released from rest near a large charged conducting sheet. Why does the particle initially accelerate toward or away from the sheet?

TopperSquare Expert · Accepted Answer

A large charged conducting sheet creates a uniform electric field near its surface ($ E = \frac{\sigma}{\varepsilon_0} $). If the sheet is positively charged, the field points away from it; if negatively charged, it points toward it. A charged particle experiences a force $ F = q E $. Depending on the particle's charge (same or opposite to the sheet), it accelerates toward (opposite charges) or away (like charges) due to this force, as the field exerts an attractive or repulsive force.

Question 2

Three charges $ +7 \, \mu\text{C} $, $ -4 \, \mu\text{C} $, and $ +2 \, \mu\text{C} $ are at $ (0, 0, 0) $, $ (7, 0, 0) $, and $ (0, 7, 0) \, \text{m} $. What is the potential at $ (7, 7, 0) \, \text{m} $? (Take $ \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} $).

TopperSquare Expert · Accepted Answer

Distances: $ r_1 = \sqrt{7^2 + 7^2} = 7\sqrt{2} \, \text{m} $, $ r_2 = 7 \, \text{m} $, $ r_3 = 7 \, \text{m} $. $ V = 9 \times 10^9 \left( \frac{7 \times 10^{-6}}{7\sqrt{2}} + \frac{-4 \times 10^{-6}}{7} + \frac{2 \times 10^{-6}}{7} \right) $. $ V = 9 \times 10^9 \left( \frac{7 \times 10^{-6}}{9.899} - \frac{4 \times 10^{-6}}{7} + \frac{2 \times 10^{-6}}{7} \right) $. $ V = 9 \times 10^9 \left( 0.707 \times 10^{-6} - 0.571 \times 10^{-6} + 0.286 \times 10^{-6} \right) $. $ V = 9 \times 10^9 \times 0.422 \times 10^{-6} = 3798 \, \text{V} $.

Question 3

In a system of two capacitors connected in series across a battery, why does the capacitor with smaller capacitance store more energy compared to the one with larger capacitance?

TopperSquare Expert · Accepted Answer

In series, the charge $ Q $ on each capacitor is the same. Energy stored in a capacitor is $ U = \frac{Q^2}{2C} $. For a smaller capacitance $ C $, the denominator $ 2C $ is smaller, so $ U $ is larger compared to a capacitor with larger $ C $. Alternatively, since $ V = \frac{Q}{C} $, the smaller capacitor has a larger voltage $ V $, and energy $ U = \frac{1}{2} C V^2 $ increases with $ V^2 $, leading to more energy in the smaller capacitor.

NEET Physics — Electrostatic Potential and Capacitance

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NEET Physics: Electrostatic Potential and Capacitance — MCQ Test 10

Q6. An electric dipole with moment \( p = 6 \times 10^{-9} \, \text{C m} \) lies along the z-axis. What is the potential at \( (0, 0, -2) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q7. A spherical conductor of radius 10 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). What is the electric field at 25 cm from the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q8. Why does the electric field inside a uniformly charged spherical shell remain zero even if the shell is placed in an external uniform electric field?

Q9. A spherical conductor of radius 5 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). What is the potential at its surface? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q10. Why does the total energy stored in a system of capacitors connected in series decrease if one of the capacitors is removed and the battery remains connected?