Question

What is the magnitude and direction of an electric field that exerts a $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathit{N}$upward force on a $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{75}\mathbf{}\mathit{\mu }\mathit{C}\mathit{a}$ charge?

Short answer

The magnitude of the electric field is $11.43N/C$, and the direction is downward.

Step-by-step solution

Given Data

Electric field Eexerts a force $F\left(=2.00\times {10}^{-5}N\right)$on a charge $\left(=-1.75\mu C\right)$.

Electric field

The force per unit positive test charge is known as the intensity of the electric field.

Magnitude and direction of electric field

From the definition of the electric field,

$E=\frac{F}{q}$

Here, F is the force $\left(F=2.00\times {10}^{-5}N\right)$, and q is the test charge $\left(q=1.75\mu C\right)$.

Substituting all known values,

$$\begin{gathered} E=\frac{2.00\times {10}^{-5}N}{-1.75\mu C} \\ =\frac{2.00\times {10}^{-5}N}{\left(-1.75\mu C\right)\times \left({\displaystyle \frac{{10}^{-6}C}{1\mu C}}\right)} \\ =11.43N/C \end{gathered}$$

Hence, the magnitude of the electric field is 11.43 N/C.

The negative sign of the electric field indicates that the direction of electric field is in the opposite direction of force. Since, the force is directed upward. Hence, the electric field is directed downward.