A spherical conductor of radius 15 cm has a charge of \( 6 \times 10^{-8} \, \text{C} \). What is the electric field at
**Potential due to point charge** V = (1/4π ε₀)Q/r, positive Q gives positive V, negative gives negative. At r=0.3 m from Q=3×10⁻⁸ C, V=9×10⁹×3×10⁻⁸/0.3=900 V. Work done moving charge q from infinity (V=0) to point where V=40 V is W = q V =9×10⁻⁶×40=3.6×10⁻⁴ J. For r = 0.4 m > R = 0.15 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (6 × 10⁻⁸/(0.4)²) = 9 × 10⁹ × (6 × 10⁻⁸/0.16) = 3.375 × 10³ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Electric Potential and Potential Difference
