In order to solve this question, we first have to identify the equation to solve for V. When it is a point charge, V = kQ/r. However, in this question, the charge is distributed throughout the rod, so we can't use this equation. If we consider breaking this rod into very very small pieces, we can treat each individual pieces as point charges and then sum up all potentials that results from those small pieces. This is the basic ideology on how we use this equation. dV = (k/r)*dQ.
Next, we have to alter dQ into some form of dr (or dx - they are pretty much the same thing) to perform integration that will help us solve for V. As you may know, λ (linear density) = Q/L. Hence, we can rearrange this equation into forms like these:
λL = Q
Ldλ = dQ
L*d(αx) = dQ
αL * dx = dQ (***L and x are equivalent.)
αx * dx= dQ
Now, let's think about r (distance between point A and "small pieces of rod) in terms of x. It is quite easy to note that the distance r is simply a sum of d and x. Hence, r = d + x.
Now, we can just put every information together to the original equation we started off with.
dV = (k/r)*dQ.
dV = ( k / (d + x) ) * αx * dx.
∫dV = ∫( k / (d + x) ) * αx * dx. (The limits of going to be from zero to L as that is the total length when we add up all "small pieces.")
V = ∫( (8.98755*10^9) / (4.77 + x) ) * 0.045x * dx (from 0 to 16.8)
= 3883541445 Volts.
Therefore, the potential at A is 3663541445 Volts.
Hi Sunil,
Thank you for your question.
In order to solve this question, we first have to identify the equation to solve for V. When it is a point charge, V = kQ/r. However, in this question, the charge is distributed throughout the rod, so we can't use this equation. If we consider breaking this rod into very very small pieces, we can treat each individual pieces as point charges and then sum up all potentials that results from those small pieces. This is the basic ideology on how we use this equation. dV = (k/r)*dQ.
Next, we have to alter dQ into some form of dr (or dx - they are pretty much the same thing) to perform integration that will help us solve for V. As you may know, λ (linear density) = Q/L. Hence, we can rearrange this equation into forms like these:
λL = Q
Ldλ = dQ
L*d(αx) = dQ
αL * dx = dQ (***L and x are equivalent.)
αx * dx= dQ
Now, let's think about r (distance between point A and "small pieces of rod) in terms of x. It is quite easy to note that the distance r is simply a sum of d and x. Hence, r = d + x.
Now, we can just put every information together to the original equation we started off with.
dV = (k/r)*dQ.
dV = ( k / (d + x) ) * αx * dx.
∫dV = ∫( k / (d + x) ) * αx * dx. (The limits of going to be from zero to L as that is the total length when we add up all "small pieces.")
V = ∫( (8.98755*10^9) / (4.77 + x) ) * 0.045x * dx (from 0 to 16.8)
= 3883541445 Volts.
Therefore, the potential at A is 3663541445 Volts.
Does this answer your question?