Find the potential at point A due to (a)...

Find the potential at point A due to (a) charge Q, ( b) charge on the outer surface of the conducting shell, and (c) total potential at A (Fig. 24-45).

Text Solution

Verified by Experts

It is easy to find the potential due to a point charge. But the charge on the surface of the conducting shell is nonuniform and its potential can be tricky. However, we can use the fact that the entire conductor is an equipotential surface.
Calculation: The first part, potential due to Q is easy. It can be seen that
`V_(Q ) = (q)/( 6 pi epsilon_(0) R)`
There are induced charges on outer surface of the shell As discussed before, they are nonuniformly distributed. So, we cannot calculate the potential due to a nonuniform shell at A. But let us try to calculate the potential due to this shell at C, the center of the shell. We cannot use the formula for the uniformly charged shell, but the center is equidistant from all the induced charges at the surface. Since potential is a scalar, Vat the center due to the induced charges will be
`V_( C "in" ) = (sum q_("in"))/( 4 pi epsilon_(0) R)`.
Since the shell is initially uncharged, the total charge on it will be zero. o, the potential at C due lo induced charges will be zero. Therefore, the total potential at C will be
`V_("cq") + V_("c in") = (Q) /( 8 pi epsilon_(0) R)+0.`
As we discussed before, the potential at all the points within the cavity should be the same. So, `V_(A) = V_( C)`
From this, we can say that
`V_(A) = (Q) / (Q pi epsilon_(0) R)`
`V_A = V_("AQ") + V_(" A in") = (Q)/( 8 pi epsilon_(0) R)`
`V_(" A in ") = (-Q)/( 24 pi epsilon_(0) R).`
Learn: We can see why the potential at A is negative. The induced charges near Q will be negative, and equal and opposite charges will be induced on the other side of the conductor. They will be farther off, producing a net negative potential at A.

Topper's Solved these Questions

ELECTRIC POTENTIAL
RESNICK AND HALLIDAY|Exercise SAMPLE PROBLEM 24.24 Potential due to charges on three concentric shells|1 Videos
ELECTRIC POTENTIAL
RESNICK AND HALLIDAY|Exercise SAMPLE PROBLEM 24.25 Charge and potential on a conductor that is earthed|1 Videos
ELECTRIC POTENTIAL
RESNICK AND HALLIDAY|Exercise SAMPLE PROBLEM 24.22 Conservation of mechanical energy with electric potential energy|1 Videos
ELECTRIC CHARGES AND FIELDS
RESNICK AND HALLIDAY|Exercise Practice Questions (Integer Type )|3 Videos
ELECTROMAGNETIC INDUCTION
RESNICK AND HALLIDAY|Exercise PRACTICE QUESTIONS (Integer Type)|14 Videos

Similar Questions

Explore conceptually related problems

A point charge q'q is placed at distance 'a' from the centre of an uncharged thin spherical conducting shell of radius R=2a. A point 'P' is located at a distance 4a from the centre of the conducting shell as shown. The electric potential due to induced charge on the inner surface of the conducting shell at point 'P' is

Potential at a point due to a point charge is V. The charge is doubled and also the distance of the point from the charge is doubled. The new potential is

Potential Due To A Point Charge

A solid metallic sphere of radius a is surrounded by a conducting spherical shell of radius b(bgt a ). The solid sphere is given a charge Q. A student measures the potential at the surface of the solid sphere as V and the potential at the surface of spherical shell as V_b . After taking these readings, he decides . to put charge of -4Q on the shell. He then noted the readings of the potential of solid sphere and the shell and found that the potential difference is /_\V . He then connected the outer spherical shell to the earth by a conducting wire and found that the charge on the outer surface of the shell as He then decides to remove the earthing connection from the shell and earthed the inner solid sphere. Connecting the inner sphere with the earth he observes the charge on the solid sphere as q_2 . He then wanted to check what happens if the two are connected by the conducting wire. So he removed the earthing connection and connected a conducting wire between the solid sphere and the spherical shelll. After the connections were made he found the charge on the outer shell as q_3 . Potential difference (/_\CV) measured by the student between the inner solid shere and outer shell after putting a charge -4Q is

Electric field intensity at a point B due to a point charge Q kept at a point charge Q kept at point A is 24 NC^(-1) , and electric potential at B due to the same charge is 12 JC^(-1) . Calculate the distance AB and magnitude of charge.

A solid conducting sphere of radius a is enclosed in a concentric shell of radius b gt a . The charge on the inner sphere is 3muC and on the conducting shell is -10muC . The potential difference between the conductors is V. If +10muC additional charge is given to the outer shell, the new potential difference is

RESNICK AND HALLIDAY-ELECTRIC POTENTIAL-SAMPLE PROBLEM 24.23 Finding potential on spherical conducting shell due to point charge kept outside the shell

Find the potential at point A due to (a) charge Q, ( b) charge on the ...
Text Solution
|