2)Find the potential at a distance of 0.5 m due to a charge of 10microC.
Answers
Given: Distance(r)=0.5m
Charge(q)= 10microC
To find The potential(V)
Solution: The given numerical is based on electrostatics, a branch of physics, in which we study the charges which are stationary.
From above, we have
Distance(r)=0.5m
Charge(q)= 10microC
= 10×10⁻⁶ C [∵1 microC= 10⁻⁶ C]
= 10⁻⁵C [law of indices, when the bases are the same, powers are added]
We know that, Potential(V)= (1/4π€₀)q/r, where 1/4π€₀=9×10⁹ Nm²/C² and €₀is known as the 'permittivity of free space '.
∴ V= (1/4π€₀)q/r
⇒ V=(9×10⁹)×10⁻⁵/0.5 [substituting the values of (1/4π€₀), q, r]
⇒ V= (9×10⁹⁺⁽⁻⁵⁾)/(5/10) [law of indices]
⇒ V= (9×10⁴)×10/5
⇒ V= (9×2×10⁴)
⇒ V=18×10⁴
⇒ V=1.8×10⁵ [multiplying 10 with both numerator and denominator]
Hence the required potential is 1.8×10⁵volts.