Question

Figure shows the variation of electric
field intensity E versus distance x. What
is the potential difference between the
points at x = 2 m and at x= 6m from
O?

(1) 30V
(3) 40V
(2) 60V
(4) 80V
​

Answer 1

dv =integration of E.dr

Area under E and R( distance) gives change in potential difference !!

so calculate the area between 2 to 6

Area = 30

so V = 30v

Answer 2

Answer:

The potential difference between the  points is 30 V.

(1) is correct.

Explanation:

According to figure,

The potential is

$dV=\int E\cdot dr$

Area under the curve, the graph of E and x gives the change in potential difference

Area, when x = 2

$A_{1}=\dfrac{1}{2}\times10\times2$

$A_{1}=10\ V$

Area, when x = 6

The potential is

$A_{2}=\dfrac{1}{2}(a+b)\times h$

$A_{2}=\dfrac{1}{2}(2+6)\times 10$

$A_{2}=40\ V$

The potential difference between the  points

The potential difference = Area of trapezium-Area of triangle

$V=A_{2}-A_{1}$

$V=40-10$

$V = 30\ V$

Hence, The potential difference between the  points is 30 V.