First slide

Electric field

Question

An infinite number of electric charges each equal to 5 nC (magnitude) are placed along x-axis at x =1 cm, x = 2 cm, x = 4 cm, x = 8m .......... and so on. In the setup, if the consecutive charges have opposite sign, then the electric field in newton/coulomb at x = 0 is $(\frac{1}{4 {πε}_{0}} = 9 \times 10^{9} N - m^{2} / C^{2})$

Moderate

A

$12 \times 10^{4}$
B

$24 \times 10^{4}$
C

$36 \times 10^{4}$
D

$48 \times 10^{4}$

Solution

$E = \frac{1}{4 {πε}_{0}} \cdot [\frac{5 \times 10^{- 9}}{{(1 \times 10^{- 2})}^{2}} - \frac{5 \times 10^{- 9}}{{(2 \times 10^{- 2})}^{2}} + \frac{5 \times 10^{- 9}}{{(4 \times 10^{- 2})}^{2}} - \frac{(5 \times 10^{- 9})}{{(8 \times 10^{- 2})}^{2}} + \dots . .] \Rightarrow E = \frac{9 \times 10^{9} \times 5 \times 10^{- 9}}{10^{- 4}} [1 - \frac{1}{(2)^{2}} + \frac{1}{(4)^{2}} - \frac{1}{(8)^{2}} + \dots] \Rightarrow E = 45 \times 10^{4} [1 + \frac{1}{(4)^{2}} + \frac{1}{(16)^{2}} + \dots] - 45 \times 10^{4} [\frac{1}{(2)^{2}} + \frac{1}{(8)^{2}} + \frac{1}{(32)^{2}} + \dots] \Rightarrow E = 45 \times 10^{4} [\frac{1}{1 - \frac{1}{16}}] - \frac{45 \times 10^{4}}{(2)^{2}} [1 + \frac{1}{4^{2}} + \frac{1}{(16)^{2}} + . .] E = 48 \times 10^{4} - 12 \times 10^{4} = 36 \times 10^{4} N / C$

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