Question

a sphere of radius 10 cm carries a charge of 1 uc , calculate the electric field ata distant of 30 cm from sphere​

Answer 1

Repulsive force of magnitude 6 × 10−3 N

Charge on the first sphere, q1 = 2 × 10−7 C

Charge on the second sphere, q2 = 3 × 10−7 C

Distance between the spheres, r = 30 cm = 0.3 m

Electrostatic force between the spheres is given by the relation:

F = (1/4πε0). (q1q2)/ (r2)

Where, ε0 = Permittivity of free space and (1/4πε0) =9 × 109 Nm2C−2

Therefore, force F = (9 × 109 × 2 × 10−7)/ ((0.3)2)

= 6 × 10−3N

Hence, force between the two small charged spheres is 6 × 10−3 N. The charges are of same nature. Hence, force between them will be repulsive.

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Answer 2

The electric filed at a distance of 30cm from sphere $10^{5}\frac{N}{C}$.

Explanation:

$q=1uc=1*10^{-6}$

$R=10cm$

$E_{0}=8.85*10^{-12}\frac{C^{2}}{Nm^{2}}$

$\frac{1}{4\pi E_{0}} =9*10^{9}\frac{Nm^{2}}{C^{2}}$

$E=\frac{1}{4\pi E_{0} }\frac{q}{r^{2}} }$

$r=30cm=0.3m$

$E=\frac{9*10^{9}*10^{-6}}{0.3^{2}}$

$=\frac{9*10^{9-6}}{9*10^{-2}}$

$=10^{5}\frac{N}{C}$.