Question

20 positively charged particles are kept fixed on the X-axis at points x=1m, 2m, 3m, … 20m. The first particle has a charge $1.0 \times {10^{ - 6}} C$, the second $8.0 \times {10^{ - 6}} C$, the third $27.0 \times {10^{ - 6}} C$ and so on. Find the magnitude of the electric force acting on a 1C charge placed at the origin.

Answer 1

In this question the concept of electric force is tested. With the electric force we have to calculate the summation of all the force exerted by all the particles on the 1C charge on the origin.

Complete step by step answer:
We have to calculate the electric field between the charge on the origin.
The electrostatic force between two charge ${Q_1}$and ${Q_2}$, with the separation of $r$ is given by:
$F = \dfrac{1}{{4\pi {\varepsilon _ \circ }}} \times \dfrac{{{Q_1} \times {Q_2}}}{{{r^2}}}$, where the value of the $\dfrac{1}{{4\pi {\varepsilon _ \circ }}}$is $9 \times {10^9}N{m^2}{C^{ - 2}}$.
As we can say that if a particle is denoted by n then the charge it possesses is equal to cube of n times ${10^{ - 6}}$C and the distance from the particle of 1C is n units.
Now if we understand the above concept, the magnitude of electric force acting on the 1C charge is the summation of all the forces acting on the 1C charge.
If we apply the concept and form the equation we get,
$F = \sum\limits_{n = 1}^{20} {\,\,\dfrac{1}{{4\pi {\varepsilon _ \circ }}}} \times \dfrac{{{n^3} \times {{10}^{ - 6}} \times 1}}{{{n^2}}}$
Here we can see that we have put ${n^3} \times {10^{ - 6}}$ in place of ${Q_1}$and 1 in place of ${Q_2}$, and in place of $r$we have put n.
Now if we manipulate the above equation to solve for the force, we will get the force acting on the 1C charge.
First, we will take out all the constants, out of the summation: -
$F = \dfrac{{{{10}^{ - 6}}}}{{4\pi {\varepsilon _ \circ }}}\sum\limits_{n = 1}^{20} {\,\,} \times \dfrac{{{n^3}}}{{{n^2}}}$
Now with the above equation in hand we can easily cancel out ${n^2}$, and we will get:
$F = \dfrac{{{{10}^{ - 6}}}}{{4\pi {\varepsilon _ \circ }}}\sum\limits_{n = 1}^{20} {\,\,} n$
Now we can apply the formula of summation up to n numbers, which is,
$\sum\limits_{n = 1}^{20} n = \dfrac{{n(n + 1)}}{2}$.
We have the value of n as 20, and when we put the value in the equation,
$\Rightarrow F = \dfrac{{{{10}^{ - 6}}}}{{4\pi {\varepsilon _ \circ }}} \times \dfrac{{n(n + 1)}}{2}\\\ \Rightarrow F = \dfrac{{{{10}^{ - 6}}}}{{4\pi {\varepsilon _ \circ }}} \times \dfrac{{20 \times 21}}{2}\\\ \Rightarrow F = \dfrac{{{{10}^{ - 6}}}}{{4\pi {\varepsilon _ \circ }}} \times 10 \times 21$
As we have discussed earlier that the value of the $\dfrac{1}{{4\pi {\varepsilon _ \circ }}}$is $9 \times {10^9}N{m^2}{C^{ - 2}}$.
We put this value in the equation:
$\Rightarrow F = {10^{ - 6}} \times 10 \times 21 \times 9 \times {10^9}\\\ \therefore F = \,1.89 \times {10^6}\,N$
Hence the force exerted by all the particles to the 1C particle is $1.89 \times {10^6}\,N$.

Note: During the calculation of the forces, we should consider applying the summation concept and the concept of arithmetic progression and the concept of calculating the general term. This question can be easily solved when the concept is understood by the students.