.Answer:
The value of the work done is [tex]\bf{ 5.29 qA_{0}}[/tex].
Explanation:
When a charged particle having charge [tex]q[/tex] is moving through an electric field [tex]E[/tex], the net force ([tex]F[/tex]) on the charge is
[tex]F = qE~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
and the work done ([tex]W[/tex]) by the particle is
[tex]W = \int\limits^x_0 {F} \, dx ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Given, [tex]E = \dfrac{A_{0}}{x^{1/2}}[/tex].
Substitute the value of electric field in equation (1) and then substitute the result in equation (2).
[tex]W &=& \int\limits^7_0 {q\dfrac{A_{0}}{x^{1/2}}} \, dx \\&=& qA_{0} \int\limits^7_0 {x^{-1/2}} \, dx \\&=& 2qA_{0}[x^{1/2}]_{0}^{7}\\&=& 5.29 qA_{0}[/tex]
