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(A) The density of electric field lines is directly proportional to the magnitude of the electric field.Given that the separation of field lines on th

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$4 	imes 10^{-4} \ N$

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(A) The density of electric field lines is directly proportional to the magnitude of the electric field.Given that the separation of field lines on the left (at point $B$) is twice the separation of those on the right (at point $A$),the electric field at $B$ is half the electric field at $A$.$E_B = \frac{E_A}{2} = \frac{40 \ Vm^{-1}}{2} = 20 \ Vm^{-1}$.The force $F$ on a charge $q$ in an electric field $E$ is given by $F = qE$.Given $q = 20 \ \mu C = 20 	imes 10^{-6} \ C$.$F = (20 	imes 10^{-6} \ C) 	imes (20 \ Vm^{-1}) = 400 	imes 10^{-6} \ N = 4 	imes 10^{-4} \ N$.

The electric field lines on the left have twice the separation of those on the right as shown in the figure. If the magnitude of the electric field at $A$ is $40 \ Vm^{-1}$,what is the force on a $20 \ \mu C$ charge kept at $B$?

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