Question

How much work is done on a proton to move it from the negative plate to a positive plate 4.3 cm
away if the field is 125 N/C?
(a) 5.5 x 10-23J (b) 1.1 x 10-16J (c) 8.6 x 10-19 (d) 5.4 J
​

Answer 1

Answer:

Work done, $W=8.6\times 10^{-19]\ J$

Explanation:

It is given that,

The charge on the proton, $q=1.6\times 10^{-19}\ C$

The proton is moved from negative plate to a positive plate 4.3 cm  away, d = 4.3 cm

Electric field, E = 125 N/C

Let W is the work is done on a proton. It is given by :

$W=q\times \Delta V$

$W=q\times Ed$

$W=1.6\times 10^{-19}\times 125\times 4.3\times 10^{-2}$

$W=8.6\times 10^{-19]\ J$

So, the work done on a proton is $8.6\times 10^{-19]\ J$. Hence, this is the required solution.