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 sphere of radius r is charged uniformly.at what distance from it surface electric potential is half of that of at its centre. ans:r/3

a solid sphere of radius r is charged uniformly.at what distance from it surface electric potential is half of that of at its centre.


ans:r/3

Grade:12

2 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear shefali

we know potential outside the sphere is V1 = kQ/x     where x is the distance from the center x>r

 and potential inside the sphere V1 = kQ/2r [3-x2/r2]

 for center put x=0  Vo = 3kQ/2r

condition given is V1 = Vo/2

                      kQ/x  = (3kQ/2r)/2

                         x=4r/3

 distance from the surface is  =x-r

                                         = 4r/3 -r =r/3

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed  solution very  quickly.

 We are all IITians and here to help you in your IIT JEE preparation.

All the best.
 
Regards,
Askiitians Experts
Badiuddin

Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Hello student
For a spherically charged body with radius, R
The electic potential inside the sphere at its center is given by, V = 3kQ/2R
The electric potential at a distance r from the center of the sphere( r > R), Vr = kQ/r
 
Given, Vr = V/2
kQ/r = 3kQ/4R
hence, r = 4R/3
 
Hence distance from surface of sphere = r – R
                                                               = R/3
 
Regards,
Kushagra

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