Question

the absolute electric potential at a point distance 20cm from a charge of 2 microC​

Answer 1

Answer:

no you are wrong

this can't be solved in this way because in your equation where is distance factor

Answer 2

The electric potential from 2 micro-coulomb at a distance of 20 cm is $9X10^{4} V$.

Given,

Distance from the point charge=20 cm

Value of point charge=2 micro-coulomb.

To find,

the absolute electric potential from 20 cm from 2 micro-coulomb.

Solution:

• Electric potential is the amount of work needed to move a unit charge from a reference point to a specific point against an electric field.
• It is a scalar quantity.

• It is represented by the symbol V.
• Electric potential due a point charge q at a distance r from it is given by:
• $V=\frac{kq}{r}$.
• where, K-electrostatic constant, q-charge(with sign) and r-distance from the charge.
• $k=9X10^{9} Nm^{2} /C^{2}$.
• $1 coulomb=10^{6} micro-coulombs$.
• $1 micro-coulomb=10^{-6} coulombs$.
• $2 micro-coulomb=2X10^{-6} coulombs$.

• $1 m=100 cm$.
• $1 cm=10^{-2} m$.

The electric potential will be,

$V=\frac{kq}{r}\\V=\frac{9X10^{9}X2X10^{-6}}{20X10^{-2} }\\V=\frac{18X10^{3} }{2X10^{-1} } \\V=9X10^{4} V.$

The electric potential is $9X10^{4} V$.

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