Question

The potential at a point A in an electric field is 300 volt and at a point B it is 1200 volt. What work must be done to move a positive charge of 3x10^-8 coulomb from A to B?

Answer 1

VA = 300V

VB = 1200V

q = 3 x 10^-8C

Work done = VB - VA / q = 1200 - 300 / 3 x 10^-8 = 3 x 10^10 J

Answer 2

The work must be done move a positive charge of $3\times 10^{-8}$ coulomb from A to B is $2.7\times 10^{-5 } J$.

Explanation:

Given:

The potential at a point A in an electric field is 300 volt and at a point B it is 1200 volt.

To move a positive charge of $3\times 10^{-8}$ coulomb from A to B.

To Find:

The work must be done move a positive charge of $3\times 10^{-8}$ coulomb from A to B.

Solution:

As given,the potential at a point A in an electric field is 300 volt and at a point B it is 1200 volt.

The potential at a point A in an electric field $V_{A} =300\ Volt$

The potential at a point B in an electric field $V_{B} =1200\ Volt$

The potential difference between point A and B,  $V=V_{B} - V_{A}$

                                                                                $=1200-300=900\Volt$

As given,to move a positive charge of $3\times 10^{-8}$ coulomb from A to B.

A positive charge $Q=3\times 10^{-8} coulomb$

The work must be done move a positive charge of $3\times 10^{-8}$ coulomb from A to B $=Q\times V$

                   $=3\times 10^{-8}\times 900$

                   $=2700\times 10^{-8}$

                  $=2.7\times 10^{-5 } J$

Thus, the work must be done move a positive charge of $3\times 10^{-8}$ coulomb from A to B is $2.7\times 10^{-5 } J$.

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