SATHEE: Electrostatics 3 Question 2

Electrostatics 3 Question 2

2. A solid conducting sphere, having a charge $Q$ , is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be $V$ . If the shell is now given a charge of $- 4 Q$ , the new potential difference between the same two surfaces is

(a) $- 2 V$

(b) $2 V$

(d) $V$

Show Answer

Answer:

Correct Answer: 2. (c)

Solution:

Initially when uncharged shell encloses charge $Q$ , charge distribution due to induction will be as shown,

The potential on surface of inner shell is

$V_{A} = \frac{k Q}{a} + \frac{k (- Q)}{b} + \frac{k Q}{b}$

where, $k$ =proportionality constant.

Potential on surface of outer shell is

$V_{B} = \frac{k Q}{b} + \frac{k (- Q)}{b} + \frac{k Q}{b}$

Then, potential difference is

$Δ V_{A B} = V_{A} - V_{B} = k Q \frac{1}{a} - \frac{1}{b}$

Given,

$Δ V_{A B} = V$

So, $k Q \frac{1}{a} - \frac{1}{b} = V$

Finally after giving charge $- 4 Q$ to outer shell, potential difference will be

$\begin{aligned} Δ V_{A B} = V_{A} - V_{B} & = \frac{k Q}{a} + \frac{k (- 4 Q)}{b} - \frac{k Q}{b} + \frac{k (- 4 Q)}{b} \\ = k Q \frac{1}{a} - \frac{1}{b} = V [from Eq. \end{aligned}$

Hence, we obtain that potential difference does not depend on the charge of outer sphere, hence potential difference remains same.

पिछला

अगला

Electrostatics 3 Question 2

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Electrostatics 3 Question 2

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Menu