Question

Two point charges of 1μC and -1μC are placed at x= 0 and x = 2 cm .

Answer 1

Answer:h

Explanation:

Answer 2

The electric potential energy is -2 × 10³ V

Explanation:

The complete question is:

Two point charges of 1μC and -1μC are placed at x= 0 and x = 2 cm. respectively. The electric potential at x = 4 cm will be:

Solution:

We know that Electric Potential at a distance r due to charge q is given by

$V=k\frac{q}{r}$

Where k is Coulomb's constant $=9\times 10^9$ kgm³/s4A²

Therefore,

Electric Potential at x = 4 m due to charges at x = 0 and x = 2 m

$V=k\frac{1\times 10^{-6}}{4}+k\frac{(-1\times 10^{-6})}{2}$

$\implies V=k\times 10^{-6}(\frac{1}{4}-\frac{1}{2}$

$\implies V=-9\times 10^9\times 10^{-6}\times\frac{1}{4}$

$\implies V=-9\times 10^3\times\frac{1}{4}$

$\implies V=-2.2\times 10^3$ Volts

Hope this answer is helpful.

Know More:

Q: Two point charges of 10C each are kept at distance of 3 m in vacuum. Calculate their electrostatic potential energy.?

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