Solved Numerical Problem No.7 ( Unit-11: Electrostatics ); HSSC-II (Class-12); Physics (New Book/KPK Text Board, Peshawar) With Verified Answers

The electric field at a point due to a point charge is 26 N/C and the electric potential at that point is 13 J/C. Calculate the distance of the point from the charge and the magnitude of the charge. (Answer 0.722 ｘ 10⁻⁹ C)  

Given:

Electric Field = E = 26 N/C  

Electric potential = V = 13 J/C

∴ Coulomb's Constant = K = `\frac {1}{4ㄫε_0}` = 9 ｘ 10⁹ Nm²C⁻² 

To Find:

Distance of the point from the charge = d = ?

The magnitude of the charge = q = ?

Solution:

We have

V = E d

or 

d = `\frac {V}{E}`

putting values

d = `\frac {13 J/C}{26 N/C}`

d = 0.5 m ------------------Ans. (1)

Now to find the magnitude of charge q, we have the formula for potential 

V = k `\frac {q}{d}`

by separating the q from the formula we get

q = `\frac {Vd}{k}`

putting values

q = `\frac {13 J/c ｘ 0.5 m}{9 ｘ 10⁹ Nm²C⁻² }`

q = `\frac {6.5 Jm/C}{9 ｘ 10⁹ Nm²C⁻² }`

q = 0.722 ｘ 10⁻⁹ C  --------------Ans.(2)

or

q = 7.22 ｘ 10⁻¹⁰ C 

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