IIT JEE , AIEEE Forum - Physics , Electricity

konichiwa2x

The electric potential decreases uniformly from 120V to 80V as one moves on the X axis from x= -1cm to x= +1cm. The electric field at the orgin

(A) must be equal to 20 V/cm

(B) may be equal to 20V/cm

(C) may be greater than 20V/cm

(D) may be less than 20V/cm

Can someone please explain the correct answer? This is an H.CVerma problem (page 120 , obj II - 4) the answer is given as (B) and (C).. why cant D be correct? please explain

joyfrancis

-E = dv / dr

-E_x = V_x / x

since the field is uniformly decreasing the potential at the origin would be 100V.

.: E_x = the electric field at x=+1 in the x dir.

= V_x / _x{change in potential divided by change in x co-ordinate}

=-20 / 1 = -20V/cm(i)

Now Ey = Vy / y

= -20 / y

now while using we only take the change i.e the positive change.

.: Enet = root (400 + something>=0) = Always greater than or equal to 20V/cm

It would be equal to 20V/m when there is zero displacement along the y direction.

konichiwa2x

thanks great work..

by the way, which book did u study electrostatics from?

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