Question

If 4*10^20ev energy is required to move a charge of 0.25 C between two points then what will be the potential difference between them

Answer 1

Answer:

256 V

The work done by an external agent by carrying a unit positive test charge from one point to another point in electric field is called the potential difference between those points.

Given,

Energy req. to move a charge =

$4 \times {10}^{20}ev$

=

$64j$

(Conversion :

1 ev =

$1.6 \times {10}^{-19}joule$

$4 \times {10}^{20}ev \: =4 \times {10}^{20} 1.6 \times {10}^{-19} joule$

=

$6.4 \times {10}^{1} = 64joule$

)

Charge = 0.25 C

Potential difference =

$\frac{work}{charge}$

=

$\frac{64}{0.25}$

= 256 V

Answer 2

Answer:

256 V

Explanation:

E = 4× 10^20 eV = 64 J

{ since 1eV = 1.6 × 10^-19 J,

4 × 10^20 eV = 4 × 10^20 × 1.6 ×10^-19 J = 6.4 × 10 = 64 J }

charge = 0.25 C

potential difference = work / charge = 64 / 0.25 = 256V.

THE CORRECT ANSWER IS 256 V.