Question

An infinite number of charges each equal to q are placed along the x-axis at x=1, x=2, x=4, x= 8 and so on. Find the potential and electric field at the point x=0 due to this set of charges.
A). \[5kq,\dfrac{4kq}{3}\]
B). \[2kq,\dfrac{4kq}{3}\]
C). \[3kq,\dfrac{4kq}{3}\]
D). \[7kq,\dfrac{4kq}{3}\]

Answer 1

Hint: For solving this question the students should first identify the type of infinite series that the charges form. Evidently the charges are placed at distances that form a GP. Also, the formula for potential, electric field and sum of infinite GP would be used.

Formula used:
\[V=\dfrac{1}{4\pi {{\in }_{0}}}\sum\limits_{i=1}^{N}{\dfrac{{{q}_{i}}}{{{r}_{i}}}}\]
\[E=k\sum\limits_{i=1}^{N}{\dfrac{{{Q}_{i}}}{{{r}_{i}}^{2}}}\]
\[S=\dfrac{a}{1-r}\]

Complete step by step answer:
Given that infinite charges ‘q’ are placed at distances along the x-axis ‘r’ given by:
 \[x=1,x=2,x=4,x=8\ldots \]
To find the potential at \[x=0\] we will use the superposition principle:
\[V=\dfrac{1}{4\pi {{\in }_{0}}}\sum\limits_{i=1}^{N}{\dfrac{{{q}_{i}}}{{{r}_{i}}}}\]
On substituting the values of r, we get:
\[V=\dfrac{1}{4\pi {{\in }_{0}}}\left[ \dfrac{q}{1}+\dfrac{q}{2}+\dfrac{q}{4}+\dfrac{q}{8}+....+\dfrac{q}{N} \right]\]
\[V=\dfrac{1}{4\pi {{\in }_{0}}}\left[ \dfrac{q}{1}+\dfrac{q}{{{2}^{1}}}+\dfrac{q}{{{2}^{2}}}+\dfrac{q}{{{2}^{3}}}+....+\dfrac{q}{{{2}^{N}}} \right]\]
\[V=\dfrac{q}{4\pi {{\in }_{0}}}\left[ \dfrac{1}{1}+\dfrac{1}{{{2}^{1}}}+\dfrac{1}{{{2}^{2}}}+\dfrac{1}{{{2}^{3}}}+....+\dfrac{1}{{{2}^{N}}} \right]\]
Since the above series in the square bracket is in GP, we will solve it using the formula for sum of infinite GP:
\[S=\dfrac{a}{1-r}\]
Here a is the first term
\[a=1\]
\[r=\dfrac{1}{2}\]
\[S=\dfrac{1}{(1-{1}/{2}\;)}\]
On solving we get
\[S=2\]
Substituting it in above equation:
\[V=\dfrac{2q}{4\pi {{\in }_{0}}}\]
\[\Rightarrow V=2kq\]
Here, \[k=\dfrac{1}{4\pi {{\in }_{0}}}\]
Now, Electric field at point \[x=0\] due to infinite charges is:
\[E=k\sum\limits_{i=1}^{N}{\dfrac{{{Q}_{i}}}{{{r}_{i}}^{2}}}\]
On substituting the values of Q and r, we get:
\[E=k\left[ \dfrac{q}{{{1}^{2}}}+\dfrac{q}{{{2}^{2}}}+\dfrac{q}{{{4}^{2}}}+\dfrac{q}{{{8}^{2}}}+.... \right]\]
\[E=kq\left[ \dfrac{1}{{{1}^{2}}}+\dfrac{1}{{{2}^{2}}}+\dfrac{1}{{{2}^{4}}}+\dfrac{1}{{{2}^{6}}}+.... \right]\]
The above series in the square bracket is in GP, we will solve it using the formula for sum of infinite GP:
\[S=\dfrac{a}{1-r}\]
Here a is the first term
\[a=1\] and \[r=\dfrac{1}{{{2}^{2}}}\]
\[\therefore S=\dfrac{1}{(1-{1}/{{{2}^{2}}}\;)}\]
\[S=\dfrac{1}{(1-{1}/{4}\;)}\]
\[S=\dfrac{4}{3}\]
Substituting above we get:
\[E=\dfrac{4kq}{3}\]
Hence, the correct answer is option B. \[2kq,\dfrac{4kq}{3}\].

Note: Students must note that the electric field is a vector quantity that varies from one point to another in space. Its values depend on the position of the source charges. The direction of the electric field can be determined using the polygon law of vectors.