Question

The variation of potential with distance r from fixed point as shown in figure, electric field at R = 5m is?

Answer 1

Answer:

The electric field is 2.5 V/m.

Solution:

The figure is not shown here. But the electric field is defined as the rate of change of potential with respect to the distance from the fixed point.  

Thus, the Electric field can be determined by the formula

$\bold{E=-\frac{d V}{d r}}$

Here, V is the potential difference and r is the distance from fixed point. If there are 6 points that means the total distance is taken from 1 m to 6 m with a period of 1m then at R = 5m will lie between 4m and 6m.  

So, dr = 6 - 4 =2 m

Let us consider that potential at r = 0 is 0 and potential at r = 5 m is 5.  

Then

$\bold{E=-\frac{(0-5)}{(6-4)}=\frac{5}{2}=2.5\ \mathrm{V} / \mathrm{m}}$

Thus the electric field will be 2.5 V/m.

Answer 2

Explanation:

The electric field is 2.5 V/m.

Solution:

The figure is not shown here. But the electric field is defined as the rate of change of potential with respect to the distance from the fixed point.

Thus, the Electric field can be determined by the formula

\bold{E=-\frac{d V}{d r}}E=−

dr

dV

Here, V is the potential difference and r is the distance from fixed point. If there are 6 points that means the total distance is taken from 1 m to 6 m with a period of 1m then at R = 5m will lie between 4m and 6m.

So, dr = 6 - 4 =2 m

Let us consider that potential at r = 0 is 0 and potential at r = 5 m is 5.

Then

\bold{E=-\frac{(0-5)}{(6-4)}=\frac{5}{2}=2.5\ \mathrm{V} / \mathrm{m}}E=−

(6−4)

(0−5)

=

2

5

=2.5 V/m

Thus the electric field will be 2.5 V/m.