Question

4. Two point charges +q, & -q are kept is distance
apart in a uniform external electric field
Ē. Find the amount of work done in assembling this
System of charges.​

Answer 1

Answer :-

$\to \boxed{ \bf{ w \: (work) = \frac{k {q}^{2} }{{r}^{2}} }} \:$

To Find :-

→ Work done .

Explanation :-

According to the question

→ Point charges +q and -q are kept at a distance of "r" in a uniform Electric field Ē .

We know that ,

→ Electric field

$\to \boxed{\bf{\vec{ E } = \frac{F}{q}}}$

and, we know that the force (F) between two point charges which are placed at a distance of "r" .

$\implies \bf{ F = k \: \frac{q \times q}{ {r}^{2} } } \\ \\ \because \boxed{\sf{k = \frac{1}{4 \pi \:\epsilon_{o}}}} \\ \therefore \\ \\ \to \:\bf{\vec{ E } = \frac{ \frac{k {q}^{2} }{{r}^{2}} }{q}} \\ \\ \to \boxed{ \bf{ \vec{ E } \: = \frac{kq}{{r}^{2}}}}$

and, we know that

$\leadsto \boxed{ \bf{\vec{ E } = \frac{work(w)}{charge(q)}}}\\ \therefore \\ \\ \to \bf{ \frac{kq}{{r}^{2}} = \frac{w}{q} } \\ \\ \to \boxed{ \bf{ w = \frac{k {q}^{2} }{{r}^{2}} }}$

Answer 2

$\large \underline{ \blue{ \boxed{ \bf \green{Required \: Answer}}}}$