Question

Abcd is a square of side 5 metre and charges of + 50 - 50 + 50 kilometre acb find electric field b

Answer 1

Answer:

3k

Explanation:

Abcd is a square of side 5 metre and charges of + 50 - 50 + 50 Coulomb at  acd find electric field b

ab = 5m  bc = 5m  bd = 5√2m

Let say ab is along x axis

Charge at a =  50 C   charge at d = 1 C

repulsive force at d by A

FA  = k * 50 * 1 / 5²  = 2k  ( along +ve y axis)

Charge at b =  50 C   charge at d = 1 C

repulsive force at d by A

FB  = k * 50 * 1 / 5√2²  = k   ( 135° along x axis)

-k/√2  along  along x axis

k/√2  along  along y axis

Charge at c =  -50 C   charge at d = 1 C

Attractive force at d by C

FC  = k * 50 * 1 / 5²  = 2k  ( along +ve x axis)

Net force at D =   2k - k/√2 along x axis  & 2k + k/√2 along y axis

Net force = k√ ( (2 - 1/√2)² + (2 + 1/√2)² )

= k √ 4 + 1/2 - 4/√2 + 4 + 1/2 + 4/√2)

= k √9

= 3k

Answer 2

Answer:27 * 10^9 or (3k)

Explanation:hope it will clear your doubt