Question

b)Relation between electric field (E) and electric potential (V) with distance (r) due to a point charge (q)​

Answer 1

Answer

• V = Er

Given

• Electric field (E) and electric potential (V) with distance (r) due to a point charge (q)​

To Find

• Relation b/w

Solution

We know that ,

$\tt E=\dfrac{kq}{r^2}\;\; \& \;\;V=\dfrac{kq}{r}$

where ,

• E denotes electric field
• k denotes constant
• q denotes charge
• r denotes distance

Now , we need to find the relation b/w E , V and r due to q .

$\implies \tt E=\dfrac{kq}{r^2}\\\\\implies \tt E=\dfrac{kq}{r}.\dfrac{1}{r}\\\\\implies \tt E=V.\dfrac{1}{r}\\\\\implies \bf E=\dfrac{V}{r}\\\\\implies \bf V=Er$

Answer 2

Given :-

Ler us Write the Given Quantities from the Question.

• Electric Field ( E ) and Electric Potential ( V ) With distance ( r ) due to point Charge ( q )

To FinD :-

• The Relation Between the Quantities.

SoluTioN :-

We know that

$\boxed{ \bold{ \pink{★E = \frac{kq}{ {r}^{2} } }{} }{} }{}$

Where,

E = Electric Field

k = Constant

q = Charge

r = Distance

Now,

The Relation between E, V and r due to q

$★E = \frac{kq}{ {r}^{2} } \\ \\ E = \frac{kq}{r} \times \frac{1}{r} \\ \\ E = V \times \frac{1}{r} \\ \\ E = \frac{V}{r} \\ \\ \bold{ \red{★V= Er}{} }{}$

Hence Prooved the Relation.