Question

Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de Graaff).

Short answer

The electric field’s magnitude is $1.125\times {10}^{7}N/C$.

Step-by-step solution

Electric field

The imaginary space around a charge in which another test charge experiences some force is known as electric field. The formula for Electric field strength is ,

$E=\frac{F}{q_{\circ }}----------(1.1)$

Here,E is the strength of the electric field,F is the Coulomb force that acts on the test charge.

Magnitude of electric field

The Coulomb force that acts on the test charge is-

$F=\frac{Kq{q}_{\circ }}{r^2}----------(1.2)$

Here,K is the electrostatic force constant $\left(K=9\times {10}^{9}N-m^2/C^2\right)$,q is the charge $\left(q=5.00mC\right)$,$q_{\circ }$ is the test charge, andr is the distance (r = 2.00 m).

The expression for the electric from equation (1.1) and (1.2) is given as,

$E=\frac{Kq}{r^2}$

Substituting all known values,

$$\begin{gathered} E=\frac{\left(9\times {10}^{9}N-m^2/C^2\right)\times \left(5.00mC\right)}{\left(2.00m\right)\left(2\right)} \\ =\frac{\left(9\times {10}^{9}N-m^2/C^2\right)\times \left(5.00mC\right)\times \left({\displaystyle \frac{{10}^{-3}C}{1mC}}\right)}{{\left(2.00m\right)}^2} \\ =1.125\times {10}^{7}N/C \end{gathered}$$

Hence, the electric field at the required distance is $1.125\times {10}^{7}N/C$.