Question

Four charges are placed on corners of a square as shown in figure having side of 5 cm. If Q is one micro coulomb, then electric field intensity at centre will be

• A. $1.02 \times 10^{7}N/C$upwards
• B. $2.04 \times 10^{7}N/C$downwards
• C. $2.04 \times 10^{7}N/C$upwards
• D. $1.02 \times 10^{7}N/C$downwards

Answer 1

Answer: (A)

$1.02 \times 10^{7}N/C$upwards

Answer 2

$|E_{C} ⥂ | > |E_{A} ⥂ |$ so resultant of $E_{C}\& E_{A}$ is $E_{CA} = E_{C} - E_{A}$directed toward Q

Also $|E_{B}| > |E_{D} ⥂ |$so resultant of $E_{B}$and $E_{D}$ i.e.

$E_{BD} = E_{B} - E_{D}$ directed toward – 2Q charge hence Net electric field at centre is

$E = \sqrt{\left( E_{CA} \right)^{2} + \left( E_{BD} \right)^{2}}$ .… (i)

By proper calculations

$|E_{A}| = 9 \times 10^{9} \times \frac{10^{- 6}}{\left( \frac{5}{\sqrt{2}} \times 10^{- 2} \right)^{2}} = 0.72 \times 10^{7}N ⥂ / ⥂ C|E_{B}| = 9 \times 10^{9} \times \frac{2 \times 10^{- 6}}{\left( \frac{5}{\sqrt{2}} \times 10^{- 2} \right)^{2}} = 1.44 \times 10^{7}N ⥂ / ⥂ C$;

$|E_{C}| = 9 \times 10^{9} \times \frac{2 \times 10^{- 6}}{\left( \frac{5}{\sqrt{2}} \times 10^{- 2} \right)^{2}} = 1.44 \times 10^{7}N ⥂ / ⥂ C$

$|E_{D}| = 9 \times 10^{9} \times \frac{10^{- 6}}{\left( \frac{5}{\sqrt{2}} \times 10^{- 2} \right)^{2}} = 0.72 \times 10^{7}N ⥂ / ⥂ C$; So,

$|E_{CA}| = |E_{C}| - |E_{A}| = 0.72 \times 10^{7}N/C$and

$|E_{BD}| = |E_{B}| - |E_{D}| = 0.72 \times 10^{7}N ⥂ / ⥂ C.$ Hence from

equation – (i) $E = 1.02 \times 10^{7}N ⥂ / ⥂ C$upwards