Question

Two point-like charges are placed as shown in the figure, r1 = 18.0 cm and r2 = 52.0 cm. Find the magnitude of the electric field at point (p) shown in the figure. Let q1 = −17.0 µC, q2 = +22.0 µC.

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Answer 1

Since charge q1 is negative charge So direction of electric field E1 due to charge q1 will be towards q1 while direction of electric field E2 due to positive charge q2 will be away from q2.As shown in the figure.

In this way both field are in the same direction So net field at P-

$\mathbf{E=E_{1}+E_{2}}$    

$\Rightarrow \mathbf{E=k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{2}}{r_{2}^{2}}}$

$\Rightarrow \mathbf{E=k[\frac{q_{1}}{r_{1}^{2}}+\frac{q_{2}}{r_{2}^{2}}]}$

$\Rightarrow \mathbf{E=9\times 10^{9}[\frac{17\times 10^{-6}}{(0.18)^{2}}+\frac{22\times 10^{-6}}{(0.52)^{2}}]}$

$\Rightarrow \mathbf{E=9\times 10^{9}\times 10^{-6}[524.69+81.36]}$

$\Rightarrow \mathbf{E=5.45\times 10^{6}N/C}$

Direction of net field is towards the negative charge.

Note:-In the formula we should put only magnitude of charge and not with the -ve sign.Because sign is used to find the direction.