Question

Three point charges of $3\times {10}^{-9}C$ are situated at each of three corners of a square, whose side is $15cm$. The magnitude and direction of the electric field at the vacant corner of the square are:

• A. $2130V/m$ along the diagonal
• B. $9622V/m$ along the diagonal
• C. $22.0V/m$ along the diagonal
• D. $zero$

Answer 1

The correct option is A $2130V/m$ along the diagonal

$E_{AB}^{\gamma }=\frac{KQ}{r^2}=\frac{9\times {10}^9\times 3\times {10}^{-9}}{15\times {10}^{-2}\times 15\times {10}^{-2}}=\frac{27\times {10}^4}{225}=1.2\times {10}^3$ $N/C$

$E_{AC}^{\gamma }=\frac{KQ}{r^2}=\frac{9\times {10}^9\times 3\times {10}^{-9}}{15\times {10}^{-2}\times 15\times {10}^{-2}}=\frac{27\times {10}^4}{225}=1.2\times {10}^3$ $N/C$

We can observe from the figure that $E_{AB}^{\gamma }\perp E_{AC}^{\gamma }$

Using parallelogram law of vector addition,

$E_{r1}^{\gamma }=\sqrt{(E_{AB}{)}^2+(E_{AC}{)}^2}=1.7\times {10}^3$ $N/C$ direction along the diagonal.

For, $E_{AC}^{\gamma }=\frac{KQ}{(r^{\prime }{)}^2}=\frac{9\times {10}^9\times 3\times {10}^{-9}}{(21\times {10}^{-2}{)}^2}=\frac{6.1\times {10}^{-2}}{{10}^{-4}}=6.1\times {10}^2=0.61\times {10}^3$ $N/C$

Directed along the diagonal.

Therefore $E_{AC}^{\gamma }\parallel E_{r1}^{\gamma }$, resultant $E_r^{\gamma }$ will be given by,

$E^{\gamma }=2.31\times {10}^3$ $N/C=2130$ $V/m$ directed along the diagonal of the square.