Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field: (a) at the mid-point of the line joining the two charges, and (b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.
Solution:
(a) At mid point C of line joining two charges, electric potential is VC = CCA+VCB or VC = 4πε01[rCAqA+rCBqB] = 9 × 109[151.5+152.5] × 10−210−16 or VC = 2.4 × 105 V and electric field at point C is EC = ECB−ECA [As ECB > ECA and they are directed opposite to each other] or EC = 4πε01[rCB2qB+rCA2qA] = 9 × 109[1522.5−1521.5] × 10−4m210−6C or EC = 4 × 105Vm−1 directed in direction of ECB→ i.e. from C to A. (b) At given point P on perpendicular bisector of line joining two charges, electric potential is VP = VPA+VPB or VP = 4πε01[rPAqA+rPBqB] = 9 × 109[181.5+182.5] × 10−2m10−6C or VP = 2.0 × 105 V and horizontal component of net electric field at point P is Ex = EPA cos θ - EPB cos θ = (EPA−EPB) cos θ or Ex = 4πε01[rPA2qA+rPB2qB] cos θ = 9 × 109[(181.5)2(182.5)2] × 10−410−6×1815 or Ex = - 2.3 × 105Vm−1 whereas vertical component of net electric field at point p is Ey = EPA sin θ + EPB sin θ = [EPA+EPB] sin θ or Ey = 4πε01[rPA2qA+rPB2qB] sin θ = 9 × 109[(181.5)2+(182.5)2] × 10−410−16×1810 or Ey = 6.2 × 105Vm−1 So, magnitude of net electric field at point P is EP = Ex2+Ey2 = 2.32+6.22 × 105 or EP = 6.6 × 105Vm−1 directed at an angle tan θ = ExEy = −2.3×1056.2×105 = - 2.6956 or θ = - 69.6° with the horizontal in –ve x-direction i.e. at 69.6° with BA.