Question

A positive charge has magnitude 2.25×10^-10 coulomb . calculate electric potential a distance of 4cm from charge​

Answer 1

Answer:

Find the magnitude of an isolated positive point charge, if the electric potential at 10

Answer 2

Explanation:

Given

magnitude of charge particle(q) = 2.25 × 10^-10 C

Distance of charge particle from point charge

= 4 cm = 4× 10^-2 m

$potiential \: (v) \: = \frac{q}{4\pi\epsilon \: r} \\ \\ = \frac{2.25 \times {10}^{ - 10} }{4\pi\epsilon \: 4 \times {10}^{ - 2} } \\ \\ (\epsilon \: = \: 8.85 \times {10}^{ - 12} ) \\ \\ = \frac{2.25 \times {10}^{ - 10} }{4\pi \times 8.85 \times {10}^{ - 12} \times 4 \times {10}^{ - 2} } \\ \\ = 50.579 \: volt$