Physics, asked by gauribodkhe63, 2 months ago

At a point A, there is an electric field of 400V/m

and potential difference of 2000 V. Then the

distance between the point charge and A is​

Answers

Answered by anubhavkumar08021999
0

Explanation:

Here, the electric field,

E

=Ecos45

i

^

+Esin45

j

^

=

2

E

(

i

^

+

j

^

)

now, V

A

−V

B

=∫

A

B

E

.

dr

=∫

A

B

2

E

(

i

^

+

j

^

).(dx

i

^

+dy

j

^

)=

2

E

[∫

0

0.03

dx+∫

0.02

0

dy]

or V

A

−V

B

=

2

400

[0.03−0.02]=2.8V

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