Question

the electric field due to a point charge is 20N/C and the electric potential at that point is 10j/C. Calculate the distance of the point from the charge and also find the magnitude of the charge.

Answer 1

The magnitude of the charge is 0.56 nC

Given:  

Electric field = 20 N/C

Electric potential = 10 J/C

Solution:  

The distance from the point of charge can be calculated through the formula of electric field and electric potential.

Electric field, $E=\frac{k \times q}{r^{2}}$

Potential difference, $V=\frac{k q}{r}$

On dividing both the formula, we get,

$\Rightarrow \frac{V}{E}=r$

$r=\frac{10}{20}$

$\Rightarrow r=0.5 \mathrm{m}$

So, the distance from the point from the charge is  0.5 m

Now, magnitude of charge,

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

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

$q=\frac{10 \times 0.5}{9 \times 10^{9}}$

$q=0.56 \times 10^{-9}$$q = 0.56 \ nC$

So, the magnitude of the charge is 0.56 nC

Answer 2

Answer:

The electric field due to a charge at a point is 20N/C and the potential at that point is 10J/C, so the distance between the charge and point is 0.5m and the charge is $0.55*10^{-9}C$

Explanation:

• The electric field due to a point charge(q) at a distance (r) is given by the formula $E=Kq/r^{2}$

• The potential due to the charge at that point is $V=Kq/r$

        where the constant $K=9*10^{9}$

From the question, we have

the magnitude of the electric field $E=20N/C$

the electric potential $V=10J/C$

$V=Kq/r=10$

$E=Kq/r^{2}=40$

take the ratio of potential to field

⇒ $V/E=r=10/20=0.5m$

now $Kq/r=\frac{9*10^{9}*q }{0.5}=10$

⇒ $q=\frac{5}{9*10^{9} }=0.55*10^{-9}C$