Question

llustration 9. Calculate the electric field at origin due to infinite number of charges as shown in figures (a) and (b) below.​

Answer 1

Answer:

figure (a) : E=2kq

figure (b) : E=2/3kq

Explanation:

Formula Used

$E=\frac{kq}{r^{2}}$

Figure (a)

$E = \frac{kq}{1^{2} } + \frac{kq}{2^{2} } + \frac{kq}{4^{2} } + ....\\E = kq [\frac{1}{1^{2} } +\frac{1}{2^{2} } + \frac{1}{4^{2} }+ ....$                              ...1

Here we need to use sum of an infinite gp formula,

S∞ = $\frac{a}{1-r}$

Here, a=1, r=2

S∞ = $\frac{1}{1-\frac{1}{2} }$ = 2

Substituting in equation 1, we get

E= 2kq

Figure b

$E = kq [ \frac{1}{1^{2} } - \frac{1}{2^{2} } + \frac{1}{4^{2} } -...$

Grouping positive terms together and negative terms together, we get

$E= kq[(\frac{1}{1^{2} } + \frac{1}{4^{2} } + ....) - ( \frac{1}{2^{2} } + \frac{1}{8^{2} } +...)]$                ....1

Now we are to consider two gps.

I. a=1, r=1/4

  S∞ = $\frac{1}{1-\frac{1}{4} }$ = 4/3

II. a=1/2, r=1/4

   S∞ = $\frac{\frac{1}{2} }{1-\frac{1}{4} }$ = 2/3

Substituting in equation 1

E= kq{$\frac{4}{3} - \frac{2}{3}$}

E= 2/3kq