Question

Potential due to system of charges​

Answer 1

Answer:

$\large\bold\red{\frac{7kQ}{4r}}$

Explanation:

First of all , Let's know what Potential is ?

• Potential (V) :- At any point in an electric field, it's the amount of work done by an external agent against electric forces in moving a unit positive charge from Infinity to that point Without changing it's Kinetic Energy.

Therefore,

The potential due to a Continuous system of charges is given by,

• $\large \boxed{\large \bold \red{\int k \frac{dq}{r} }}$

Now,

We know that,

• Potential due to system of charges at a point is the sum of individual potential by each charge at that point.

Therefore,

In the given question,

We have,

$= > V = \frac{kQ}{r} + \frac{kQ}{2r} + \frac{kQ}{4r} \\ \\ = > V = kQ( \frac{1}{r} + \frac{1}{2r} + \frac{1}{4r} ) \\ \\ = > V = kQ( \frac{4 + 2 + 1}{4r} ) \\ \\ = > \large\bold{ V = \frac{7kQ}{4r}}$

Answer 2

$\huge{ \underline{\underline{ \fcolorbox{white}{pink}{\sf{Answer :-}}}}}$

First of all , Let's know what Potential is ?

Potential :- At any point in an electric field, it's the amount of work done by an external agent against electric forces in moving a unit positive charge from Infinity to that point Without changing it's Kinetic Energy.

Therefore,

The potential due to a Continuous system of charges is given by,

$\: \large \boxed{\large \bold \pink{\int k \frac{dq}{r} }} </p><p>$

Now,

For different system of charges,

Potential is also different.