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Charges $+ 9Q$ and $- 4Q$ are placed as shown in the figure. The point at which electric field intensity is zero at a distance from B in the line joining AB will be:

Last updated date: 14th Jun 2024
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Hint:To solve this question, we will first assume the point at which electric field intensity is zero at a distance from B. Then after applying the electric field formula for both charges and their respective distance, we will get our required answer, i.e., the distance which we have assumed by making one quadratic equation.

We have been given a figure where two charges $+ 9Q$ and $- 4Q$ are placed. We need to find the point at which electric field intensity is zero at a distance from B in the line joining AB.Let x be the distance from B where electric field is zero, then the distance from A to that point will be $\left( {l{\text{ }} - {\text{ }}x} \right).$ Now let us see the figure mentioned below.

We know that, Electric field due to charge Q at a distance r is given by,
$\Rightarrow E = k\dfrac{Q}{{{r^2}}}$
So, on putting the values in the above formula, we get
The electric field due to $+ 9Q$ charge, ${E_1} = k\dfrac{{9Q}}{{{x^2}}}$
And the electric field due to $- 4Q$ charge, ${E_2} = k\dfrac{{ - 4Q}}{{{{\left( {l - x} \right)}^2}}}$
Since, these fields are in the same direction, Therefore, the net field will be zero.
Thus, ${E_1} + {E_2} = 0$
$\Rightarrow k\dfrac{{9Q}}{{{x^2}}} - k\dfrac{{ - 4Q}}{{{{\left( {l - x} \right)}^2}}} = 0$
$\Rightarrow\dfrac{9}{{{x^2}}} - \dfrac{4}{{{{\left( {l - x} \right)}^2}}} = 0$
$\Rightarrow 9{\left( {l - x} \right)^2} - 4{x^2} = 0 \\ \Rightarrow{\left( {3l - 3x} \right)^2} - {\left( {2x} \right)^2} = 0 \\ \Rightarrow \left( {3l - 3x + 2x} \right)\left( {3l - 3x - 2x} \right) = 0 \\ \Rightarrow x = 3l,x = \dfrac{{3l}}{5} \\$

Hence, the point at which electric field intensity is zero is at a distance from B in the line joining AB will be $3l.$

Note: Students should note, that we have only considered the value of x as $3l,$ because distance can be considered only one, two distance value is not possible here, and we have ignored the fraction value, because it will be easier for us, if we only consider a number which is not in a fraction.