Question

A charge of $$1\mu C$$ is divided into two parts such that their charges are in the ratio of 2 : 3. These two charges are kept at a distance 1 m apart in a vacuum. Then, the electric force between them (in N) is:
A. 0.216
B. 0.00216
C. 0.0216
D. 2.16

Answer 1

Hint: The formula for calculating the electrostatic force between the two charges should be used to find the electric force between the two charges. The ratio should be used as a fraction to multiply with the charge given to find the value of two charges.

Formula used:
$$F={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{q_1{q}_2}{d^2}}$$

Complete step by step answer:
From given, we have the data,
The amount of the charge is, $$q=1\mu C$$
The ratio of the charge is divided into 2 parts = 2 : 3
The distance between the charges, d = 1 m

First, let us compute the values of two charges, as a charge is being divided into the 2 parts with the ratio 2 : 3, let the 2 charges be $$q_1$$and $$q_2$$.
So, we have,
$$\begin{array}{rl}& {\displaystyle \frac{q_1}{q_2}}={\displaystyle \frac{2}{3}}\\ & \Rightarrow {\displaystyle \frac{q_1}{q_2}}={\displaystyle \frac{{}^2/{}_5}{{}^3/{}_5}}\end{array}$$
Continue the further calculation to find the value of charges.
So, we get,
$$\begin{array}{rl}& q_1={\displaystyle \frac{2}{5}}\times q\\ & \Rightarrow q_1={\displaystyle \frac{2}{5}}\times 1\mu \end{array}$$
Therefore, the value of the one of the charges is as follows.
$$q_1={\displaystyle \frac{2}{5}}\mu C$$
Now, compute the value of the other charge.
$$\begin{array}{rl}& q_2={\displaystyle \frac{3}{5}}\times q\\ & \Rightarrow q_2={\displaystyle \frac{3}{5}}\times 1\mu \end{array}$$
Therefore, the value of the one of the charges is as follows.
$$q_2={\displaystyle \frac{3}{5}}\mu C$$
As now we have obtained the values of the charges, using the formula for calculating the force, let us compute the same.
The formula for calculating the electric force between the charges is,
$$F={\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{q_1{q}_2}{d^2}}$$
Where $$q_1,q_2$$ are the charges and d is the distance between the charges.
The value of $${\displaystyle \frac{1}{4\pi {\epsilon }_0}}$$ is equal to a numerical value given by $$9\times {10}^9$$
Now substitute the given value of the distance between the charges and the obtained values of the charges in this equation.
$$\begin{array}{rl}& F=9\times {10}^9\times {\displaystyle \frac{2}{5}}\times {10}^{-6}\times {\displaystyle \frac{3}{5}}\times {10}^{-6}\times {\displaystyle \frac{1}{1^2}}\\ & \Rightarrow F=0.00216N\end{array}$$

As the value of the electric force between the two charges (in N) is 0.00216 N, thus, the option (B) is correct.

Note:
The things to be on your finger-tips for further information on solving these types of problems are: The units of the parameters should be taken care of. As for the given problem, both the charges had the same micro Coulomb unit. In some cases, they may be different like, one may be given as nano Coulomb and the other as micro Coulomb.