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# Two point charges of 5 micro coulomb and 20 micro coulomb and seperated by a distance of 2m.Find the point on the line joining them at which electric feild is zero.

Khimraj
3007 Points
4 years ago
The electric field will be zero in between the line joining two charge because field of one is cancelled by sending one whole on either side the field by one is supported by another so it will add up and cannot be zero. Let this point be at distance x from 20 micro coulomb charge then.K(5*10^-6)/(2-x)² = K(20*10^-6)/x² then 1/2-x = 2/x and it is x = 4/3 m. So at a distance 4/3m from larger charge electric field become zero.Hope it clears. If u have doubts then please clearify. And if u like my answer then please approve answer
ankit singh
one year ago
The point on the line joining the two point charges be given by,
Given: Charge q_{1}=5 \times 10^{-19} C
Charge q_{2}=20 \times 10^{-19} C
Distance between the point charges q_{1} and q_{2}  = 2 m
Let us assume the point at which electric field intensity is zero be at a distance from q_{2} = x
And from q_{1} = 2-x
As given that \bold{E_{1}-E_{2}=0}
\begin{array}{l}{E_{1}=E_{2}} \\ \\{E=\frac{K q}{r^{2}}} \\ \\{\frac{K q_{1}}{r_{1}^{2}}=\frac{K q_{2}}{r_{2}^{2}}} \\ \\{\frac{q_{1}}{r_{1}^{2}}=\frac{q_{2}}{r_{2}^{2}}}\end{array}
\begin{array}{l}{\frac{5 \times 10^{-19}}{(2-x)^{2}}=\frac{20 \times 10^{-19}}{x^{2}}} \\ \\{\frac{5}{(2-x)^{2}}=\frac{20}{x^{2}}} \\ \\{\frac{1}{(2-x)^{2}}=\frac{4}{x^{2}}} \\ \\{\text { Squaring both sides, } \frac{1}{(2-x)}=\frac{2}{x}}\end{array}
x = 4 – 2x
x + 2 x = 4
3x =4
x=\frac{4}{3}
x = 1.33  m.
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