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NEET PhysicsMedium

A spherical conductor of radius 10 cm10 \text{ cm} has a charge of 3.2×10−7 C3.2 \times 10^{-7} \text{ C} distributed uniformly. What is the magnitude of electric field at a point 15 cm15 \text{ cm} from the centre of the sphere? (14πϵ0=9×109 Nm2/C2\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2)

1

1.28×104 N/C1.28 \times 10^4 \text{ N/C}

2

1.28×105 N/C1.28 \times 10^5 \text{ N/C}

3

1.28×106 N/C1.28 \times 10^6 \text{ N/C}

4

1.28×107 N/C1.28 \times 10^7 \text{ N/C}

Step-by-Step Solution

Electric field outside a conducting sphere is E=14πϵ0Qr2E = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}. Here Q=3.2×10−7 CQ = 3.2 \times 10^{-7} \text{ C}, r=15 cm=0.15 mr = 15 \text{ cm} = 0.15 \text{ m}. E=(9×109)×3.2×10−7(0.15)2=9×3.2×1020.0225=1.28×105 N/CE = (9 \times 10^9) \times \frac{3.2 \times 10^{-7}}{(0.15)^2} = \frac{9 \times 3.2 \times 10^2}{0.0225} = 1.28 \times 10^5 \text{ N/C}.

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