Question

four charges q, -q, -2q,+2q are placed on corners of a square having side 10 cm. if q is 1 microcoulomb then what Will be electric field intensity at the centre of the square? ​

Answer 1

Answer:

4×10^-8 J/C

Explanation:

the whole math have done on the pic

Answer 2

Answer:

Concept :

Each location in space where a charge exists in any form can be regarded to have an electric field attached to it. The electric force per unit charge is another name for an electric field. E=F/Q is the formula for the electric field. E is the electric field where. Force F exists. The charge is Q. Variable magnetic fields or electric charges are frequently the source of electric fields. Volts per metre (V/m), a unit used in the SI, is used to express electric field strength. The force acting on the positive charge is assumed to be exerted in the direction of the field. The electric field is directed radially inwards toward negative point charge and radially outwards from positive charge.

Explanation:

Given that AB = BC= CD = AD =1 0 cm, q= 1×$10^{-8}$C

Therefore OA = OB = OC = OD = 5√2 × $10^{-2}$ m

The net electric field along OA, E1 = Ec - Ea

E1 = 0.18× $10^{7}$ $NC^{-1}$

The net electric field along OB, E2 = E2 - E0

E2 = 0.18× $10^{7}$$NC^{-1}$

Resultant electric field

E = √$E_{1} ^{2}$+$E_{2} ^{2}$

E = √(0.18×$10^{7}$)^2+(0.18×$10^{7}$)^2

E = 0.254 × $10^{7}$ $NC^{-1}$

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