Solved Problem on Electric Field

A point charge of 10 nC is located in an electric field. The field exerts a force of 1×10−2 N, horizontally and from left to right. Find:
a) The magnitude and direction of the electric field;
b) If the charge were replaced by a charge of −2 µC. What would be the magnitude and direction of the new force acting on the charge?

Problem data:
  • Electric charge:    q = 10 nC = 10×10−9 C;
  • Electric force:    F = 1×10−2 N.

a) Since the charge is positive, the electric force vector, \( \vec F\), and the electric field vector, \( \vec E\), are in the same direction (Figure 1).
The electric force and the electric field are related by the expression
\[ \begin{gather} \bbox[#99CCFF,10px] {\vec F=q\vec E} \end{gather} \]
Figure 1

the magnitude of the electric field can be written as
\[ \begin{gather} E=\frac{F}{q}\\[5pt] E=\frac{1\times 10^{-2}\;\mathrm N}{10\times 10^{-9}\;\mathrm C}\\[5pt] E=1\times 10^{-2}\times 10^8\;\mathrm{N/C}\\[5pt] E=1\times 10^6\;\mathrm{N/C} \end{gather} \]
The electric field will be (Figure 1)

Magnitude:  1 × 106 N/C;
Direction:  horizontal pointing to the right as the electric field.

b) For a charge of −2×10−6 C, we have an electric force of magnitude
\[ \begin{gather} F=qE\\[5pt] F=\left(-2\times 10^{-6}\;\mathrm{\cancel C}\right)\left(1\times 10^6\;\mathrm{\frac{N}{\cancel C}}\right)\\[5pt] F=-2\;\mathrm N \end{gather} \]
Figure 2

The electric force will be (Figure 2)

Magnitude:  − 2 N;
Direction:  horizontal pointing to the left, opposite to the electric field.