Question

An infinite number of charges each equal to q, are placed along the x-axis of $x = 1,x = 2,x = 4,x = 8,x = 16$ and so on. The electric field at origin O due to this set of charges is $\left( {k = \dfrac{1}{{4\pi {\varepsilon _0}}}} \right)$.



A. $kq$
B. $\dfrac{4}{3}kq$
B. $\dfrac{3}{4}kq$
C. $\dfrac{4}{9}kq$

Answer 1

Hint: The magnitude of the electric field ( E) produced by a point charge with a magnitude Q charge at a point distant from the point charge, is given by the equation $E = \dfrac{{kq}}{{{r^2}}}$, where k is a constant with an estimation of $8.99 \times {10^9}N{m^2}/{C^2}$. We will create a geometric progression as done in the solution below-

Formula used: $E = \dfrac{{kq}}{{{r^2}}}$, $\dfrac{a}{{1 - r}}$

Complete Step-by-Step solution:
 As we already know that-
$ \Rightarrow E = \dfrac{{kq}}{{{r^2}}}$
Now, electric field at 0 will be-
$
   \Rightarrow 0 = \dfrac{{kq}}{{{1^2}}} + \dfrac{{kq}}{{{2^2}}} + \dfrac{{kq}}{{{4^2}}} + \dfrac{{kq}}{{{8^2}}} + - - - - \\
    \\
   \Rightarrow 0 = kq\left( {1 + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{{\left( {{2^2}} \right)}^2}}} + \dfrac{1}{{{{\left( {{2^3}} \right)}^2}}} + - - - } \right) \\
    \\
   \Rightarrow 0 = kq\left( {1 + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^4}}} + \dfrac{1}{{{2^6}}} + - - - } \right) \\
$
Now, we know that in geometric progressions, sum to infinite is- $\dfrac{a}{{1 - r}}$. Where r is the common difference which is-
$ \Rightarrow \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{1}{4}$
Putting the values calculated in the above equation we get-
$
   \Rightarrow kq\left( {\dfrac{1}{{1 - \dfrac{1}{4}}}} \right) \\
    \\
   \Rightarrow kq\dfrac{4}{3} \\
    \\
   \Rightarrow \dfrac{4}{3}kq \\
$
Hence, option B is the correct option.

Note- The direction of the electric field created by a charge is away from the charge if the charge is positive, and toward the charge if the charge is negative. Electric field is a vector, so when there are different point charges present, the net electric field anytime is the vector total of the electric fields because of the individual charges.