Two point charges q 1 = + 4 nC and q 2 = − 9 nC are placed on the x-axis at x = 0 and x = 10 cm respectively. The position (in cm) on the x-axis where the net electric field is zero is x 0 . Fin

Question

Two point charges q 1 ​ = + 4 nC and q 2 ​ = − 9 nC are placed on the x-axis at x = 0 and x = 10 cm respectively. The position (in cm) on the x-axis where the net electric field is zero is x 0 ​ . Find the value of ∣ x 0 ​ ∣ .

TestHub · Accepted Answer

Let the point where the electric field is zero be at position $x$ . Since the charges are of opposite signs, the zero field point must lie outside the region between the charges, and closer to the smaller magnitude charge ( $q_{1}$ ). So, $x < 0$ .

The electric field due to $q_{1}$ at $x$ is $E_{1} = \frac{k ∣ q _{1} ∣}{x ^{2}}$ and due to $q_{2}$ is $E_{2} = \frac{k ∣ q _{2} ∣}{( 10 - x ) ^{2}}$ . For the net field to be zero, $E_{1} = E_{2}$ . $\frac{k ( 4 nC )}{x ^{2}} = \frac{k ( 9 nC )}{( 10 - x ) ^{2}}$ $\frac{4}{x ^{2}} = \frac{9}{( 10 - x ) ^{2}}$ Taking the square root on both sides: $\frac{2}{∣ x ∣} = \frac{3}{∣10 - x ∣}$ . Since $x < 0$ , $∣ x ∣ = - x$ and $∣10 - x ∣ = 10 - x$ . $\frac{2}{- x} = \frac{3}{10 - x}$ $2 (10 - x) = - 3 x$ $20 - 2 x = - 3 x$ $x = - 20 cm$ . Thus, $∣ x_{0} ∣ = ∣ - 20∣ = 20 cm$ .

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