
Two point charges of 20μC and 80μC are of 10 cm apart where will the electric field strength be zero on the line joining the charges from 20μC charge
(A) 0.1m
(B) 0.04m
(C) 0.033m
(D) 0.33m
Answer
567.3k+ views
Hint
As we know, the formula of electric field is $E = \dfrac{{kq}}{{{r^2}}}$. In this question we assume that at distance x from 20μC the electric field is zero then we find the total electric field at this point due to both charges and put the electric field equal to zero, we get the required result.
Complete step by step answer
It is given that, The two charges 20μC and 80μC are 10 cm apart, we have to find out the point at which the electric field is zero from the 20μC charge.
So, let the point is at the distance x from the 20μC charge then this point will be at the distance of 10-x from the 80μC charge. Then total electric field at this point due to both charges is
$ \Rightarrow E = \dfrac{{k{q_1}}}{{r_1^2}} - \dfrac{{k{q_2}}}{{r_2^2}}$
Here, r1 and r2 are the distance of the point from two charges
And $k = \dfrac{1}{{4\pi { \in _0}}}$, ∈0 is the permittivity of the free space.
Substitute the values, we get
$ \Rightarrow E = \dfrac{{20k}}{{{x^2}}} - \dfrac{{80k}}{{{{\left( {10 - x} \right)}^2}}}$
Now, as the electric field is zero then put E = 0, we get
$ \Rightarrow \dfrac{{20k}}{{{x^2}}} = \dfrac{{80k}}{{{{\left( {10 - x} \right)}^2}}}$
On solving, we get
$ \Rightarrow \dfrac{1}{2} = \dfrac{x}{{10 - x}}$
$ \Rightarrow 10 - x = 2x$
$ \Rightarrow 3x = 10$
$ \Rightarrow x = 3.3cm = 0.033m$
Hence, at the distance 0.033m from the 20μC charge the electric field is zero.
Thus, option (C) is correct.
Note
It must be noticed that the electric field is zero when the vector sum of all the fields due to all charges is zero at that point or we can say that the sum of electric field vectors has the same intensity and direction, but are opposite. Magnitude of Electric field is never negative but the direction of the electric field can be negative. It should remember that electric potential is not equal to zero even if electric field is equal to zero or vice-versa
As we know, the formula of electric field is $E = \dfrac{{kq}}{{{r^2}}}$. In this question we assume that at distance x from 20μC the electric field is zero then we find the total electric field at this point due to both charges and put the electric field equal to zero, we get the required result.
Complete step by step answer
It is given that, The two charges 20μC and 80μC are 10 cm apart, we have to find out the point at which the electric field is zero from the 20μC charge.
So, let the point is at the distance x from the 20μC charge then this point will be at the distance of 10-x from the 80μC charge. Then total electric field at this point due to both charges is
$ \Rightarrow E = \dfrac{{k{q_1}}}{{r_1^2}} - \dfrac{{k{q_2}}}{{r_2^2}}$
Here, r1 and r2 are the distance of the point from two charges
And $k = \dfrac{1}{{4\pi { \in _0}}}$, ∈0 is the permittivity of the free space.
Substitute the values, we get
$ \Rightarrow E = \dfrac{{20k}}{{{x^2}}} - \dfrac{{80k}}{{{{\left( {10 - x} \right)}^2}}}$
Now, as the electric field is zero then put E = 0, we get
$ \Rightarrow \dfrac{{20k}}{{{x^2}}} = \dfrac{{80k}}{{{{\left( {10 - x} \right)}^2}}}$
On solving, we get
$ \Rightarrow \dfrac{1}{2} = \dfrac{x}{{10 - x}}$
$ \Rightarrow 10 - x = 2x$
$ \Rightarrow 3x = 10$
$ \Rightarrow x = 3.3cm = 0.033m$
Hence, at the distance 0.033m from the 20μC charge the electric field is zero.
Thus, option (C) is correct.
Note
It must be noticed that the electric field is zero when the vector sum of all the fields due to all charges is zero at that point or we can say that the sum of electric field vectors has the same intensity and direction, but are opposite. Magnitude of Electric field is never negative but the direction of the electric field can be negative. It should remember that electric potential is not equal to zero even if electric field is equal to zero or vice-versa
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