Question

Using the symmetry of the arrangement, calculate the magnitude of the electric field in N/C at the center of the square given that qa = 96--1.00 PC and qc = 9d = +3.02 pCq. Assume that the square is 5 mon a side. qa ab 0 Oda 9c

Answer 1

Electric field is given by:

E = kq/r^2

direction of E will be towards -ve charge and away from +ve charge

Since qa is -ve, Electric field will be towards the charge in North of West direction

Since qb is -ve, Electric field will be towards the charge in North of East direction

Since qc is +ve, Electric field will be away from charge in North of east direction

Since qd is +ve, Electric field will be away from the charge in North of west direction

da C. Sinust a Eg Siny So E. Smu ET &c E. Cosus Ea. Cash + Ę. Lossy E cosure T45° 45 ad Ea. Cosus - E. Costa Ec. losus - Ed. Casus = 0 So (Enet) & Emet y = Earrin us + Eusimus + Esseny + Ed Sinus TEola El & Ela leal Ened) ya 2. Smus (ltal + Ed)

So, Now using above figure and given values we can see that from symmetry

|Ea| = |Eb| and |Ec| = |Ed|

E_net = 2*|Ea|*sin 45 deg + 2*|Ec|*sin 45 deg

E_net = (sqrt 2)*(|Ea| + |Ec|)

E_net = (sqrt 2)*(k*|qa|/ra^2 + k*|qc|/rc^2)

ra = rc = r = sqrt (a^2 + a^2)/2 = a/sqrt (2) = 5/sqrt 2

where a = side of square = 5 m

r^2 = (5/sqrt 2)^2 = 12.5

So,

E_net = (sqrt 2)*(k/r^2)*(|qa| + |qc|)

Using given values:

E_net = (sqrt 2)*(9*10^9/12.5)*(1.00*10^-6 + 3.02*10^-6)

E_net = 4093.3 N/C = 4.09*10^3 N/C

Let me know if you've any query.