Physics, asked by jamescottrell068, 4 days ago

2)Find the potential at a distance of 0.5 m due to a charge of 10microC.​

Answers

Answered by qwvilla
0

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.

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