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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 14 of 37
Marks: +1, -0
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
VCV_C = CCA+VCBC_{CA}+V_{CB}
or VCV_C = 14πε0\frac{1}{4\pi\varepsilon_0} [qArCA+qBrCB]\left[ \frac{q_A}{r_{CA}} + \frac{q_B}{r_{CB}} \right]
= 9 × 109[1.515+2.515]10^9 \left[ \frac{1.5}{15} + \frac{2.5}{15} \right] × 10−1610−2\frac{10^{-16}}{10^{-2}}
or VCV_C = 2.4 × 10510^5 V
and electric field at point C is
ECE_C = ECB−ECAE_{CB} - E_{CA} [As ECBE_{CB} > ECAE_{CA} and they are directed opposite to each other]
or ECE_C = 14πε0\frac{1}{4\pi\varepsilon_0} [qBrCB2+qArCA2]\left[ \frac{q_B}{r_{CB}^2} + \frac{q_A}{r_{CA}^2} \right] = 9 × 109[2.5152−1.5152]10^9 \left[ \frac{2.5}{15^2} - \frac{1.5}{15^2} \right] × 10−6 C10−4 m2\frac{10^{-6}\,\mathrm{C}}{10^{-4}\,\mathrm{m}^2}
or ECE_C = 4 × 105 Vm−110^5 \,\mathrm{Vm}^{-1}
directed in direction of ECB→E_{\overset{\rightarrow}{CB}} i.e. from C to A.
(b) At given point P on perpendicular bisector of line joining two charges, electric potential is
VPV_P = VPA+VPBV_{PA} + V_{PB} or VPV_P = 14πε0[qArPA+qBrPB]\frac{1}{4\pi\varepsilon_0} \left[ \frac{q_A}{r_{PA}} + \frac{q_B}{r_{PB}} \right] = 9 × 109[1.518+2.518]10^9 \left[ \frac{1.5}{18} + \frac{2.5}{18} \right] × 10−6 C10−2 m\frac{10^{-6}\,\mathrm{C}}{10^{-2}\,\mathrm{m}}
or VPV_P = 2.0 × 10510^5 V
and horizontal component of net electric field at point P is
ExE_x = EPAE_{PA} cos θ - EPBE_{PB} cos θ = (EPA−EPB)(E_{PA}-E_{PB}) cos θ
or ExE_x = 14πε0[qArPA2+qBrPB2]\frac{1}{4\pi\varepsilon_0} \left[ \frac{q_A}{r_{PA}^2} + \frac{q_B}{r_{PB}^2} \right] cos θ = 9 × 109[(1.518)2(2.518)2]10^9 \left[ \left(\frac{1.5}{18}\right)^2 \left(\frac{2.5}{18}\right)^2 \right] × 10−610−4×1518\frac{10^{-6}}{10^{-4}} \times \frac{15}{18}
or ExE_x = - 2.3 × 105 Vm−110^5 \,\mathrm{Vm}^{-1}
whereas vertical component of net electric field at point p is
EyE_y = EPAE_{PA} sin θ + EPBE_{PB} sin θ = [EPA+EPB]\left[ E_{PA} + E_{PB} \right] sin θ
or EyE_y = 14πε0[qArPA2+qBrPB2]\frac{1}{4\pi\varepsilon_0} \left[ \frac{q_A}{r_{PA}^2} + \frac{q_B}{r_{PB}^2} \right] sin θ = 9 × 109[(1.518)2+(2.518)2]10^9 \left[ \left(\frac{1.5}{18}\right)^2 + \left(\frac{2.5}{18}\right)^2 \right] × 10−1610−4×1018\frac{10^{-16}}{10^{-4}} \times \frac{10}{18}
or EyE_y = 6.2 × 105 Vm−110^5 \,\mathrm{Vm}^{-1}
So, magnitude of net electric field at point P is
EPE_P = Ex2+Ey2\sqrt{E_x^2 + E_y^2} = 2.32+6.22\sqrt{2.3^2 + 6.2^2} × 10510^5
or EPE_P = 6.6 × 105 Vm−110^5 \,\mathrm{Vm}^{-1} directed at an angle
tan θ = EyEx\frac{E_y}{E_x} = 6.2×105−2.3×105\frac{6.2 \times 10^5}{-2.3 \times 10^5} = - 2.6956 or θ = - 69.6°
with the horizontal in –ve x-direction i.e. at 69.6° with BA.
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