Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A solid conducting sphere having charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the p.d. between the surface of the solid sphere and that of the outer surface of hollow shell be V. If now shell is given a charge - 3 Q, the new p.d. between the same two surfaces is?

A solid conducting sphere having charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. Let the p.d. between the surface of the solid sphere and that of the outer surface of hollow shell be V. If now shell is given a charge - 3 Q, the new p.d. between the same two surfaces is?

Grade:12

2 Answers

Akshat Agrawal
24 Points
4 years ago
Potential differnce between solid sphere and spherical shell depends on radii of two spheres and charge on the inner sphere. Since these parameters remain the same, potential differnce under new situation remains same.
Aditya
11 Points
3 years ago
Let potential at infinity equals to 0.R-radius of shell, r-radius of sphere Now, in first case, sphere has potential kQ/r. Shell has potential 0. Therefore,pd is kQ/r. Now after giving -3Q to the shell, potential of sphere becomes( kQ/r )-(3kQ/R). Potential of shell is -3 kQ/R. Therefore pd is (kQ/r - 3kQ/r)-(-3kQ/r)=kQ/r. That is, it remains same

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free