Question

If on the x-axis electric potential decreases uniformly from 60V to 20V between x = -2m to x = +2m, then the magnitude of the electric field at the origin.
1) Must be 10 V/m
2) May be greater than 10V/m
3) Is zero
4) Is 5 V/m

Answer 1

Hint:Electric field and electric potential both are inversely proportional to the distance ‘r’. Electric field is inversely proportional to the square of the distance while potential is just inversely proportional to the distance.

Complete step by step solution:
Write the formula for electric potential:
$V = \dfrac{{kQ}}{r}$ ;
Where:
V = Electric Potential;
k = Constant;
Q = Charge;
r = Distance;
Now, write the formula for electric potential:
$E = \dfrac{{kQ}}{{{r^2}}}$ ;
E = Electric Potential;
k = Constant;
Q = Charge;
r = Distance;
Form the above two formulas for the electric potential and electric field we can deduce:
$E = \dfrac{V}{r}$ ;
Now, to find the electric field at the origin:
$ \Rightarrow dE = - \dfrac{{dV}}{{dr}}$;
$ \Rightarrow {E_{Centre}} = - \dfrac{{\left( {{V_2} - {V_1}} \right)}}{{\left( {{r_2} - {r_1}} \right)}}$;
Put in the given values in the above equation and solve:
\[{E_{Centre}} = - \dfrac{{\left( {20 - 60} \right)}}{{\left( {2 - \left( { - 2} \right)} \right)}}\];
Do the necessary calculation:
${E_{Centre}} = - \dfrac{{\left( { - 40} \right)}}{4}$;
The electric field comes out to be:
${E_{Centre}} = 10V/m$;

Now we are getting the magnitude of the electric field as 10 V/m, this value is only in the x direction, suppose the electric field might be in two directions then the value of the electric field might be greater than 10V/m.

Here both option “1” is correct. The magnitude of the electric field at the origin may be greater than 10 V/m.

Note:Here, we need to form a relation between the electric field and electric potential. First, write the formula for electric potential and then write the formula for electric field. Then put the value of the potential in the formula for the electric field and find out the magnitude of the electric field by subtracting the final potential with the initial potential.