# Electric Potential produced by Hollow Sphere

Let there be a hollow sphere (Throughout the question we will ignore the thickness of the sphere) which is positively uniformly charged,q of radius, R. Suppose there to be a test positive charge, q' at point P which is r distance away from the hollow sphere(r>R, P is outside of hollow sphere).

Now I have an doubt on evaluating Electric Potential,V at point P.

Both source charge,q and test charge,q' are +ve, so source charge will repel the test charge which will make the charge, q' to go to infinity.I have to apply external force, F' to displace the charge, q' from infinity to point P against the equal but repulsive force, F.In short, F' is displacing the q' from infinity to P.

So External Force, F' will do +ve work and Repulsive Force, F will do -ve work. This way the Electric Field,E produced by hollow sphere is Opposite to Displacement of test charge(i.e.,θ=180).

Now we know that E=k(q/(r^2))

[Please Notice that infinity is the lower limit of integration]

Also V=-∫r(E dr cosθ)=-∫r(E dr cos 180)=∫r(E dr)

At last I ended up to V=- k(q/r). I know that V=k(q/r).

I just want to know why I am having this shitty negative sign in my equation.

• – Qmechanic Jul 1 '16 at 12:22
• Sorry but I think your question is not about physics but about an error in your calculation. An answer is not of benefit to the wider community. – sammy gerbil Jul 1 '16 at 17:48

You have added the negative sign in front of your integral and then put in the cos(180) as well.

Pick one.

Since your external force is opposite in direction to the force of the sphere, and you've already put the cos(180) in there, there is no need for the extra negative sign in front.

• I am not sure this is correct. The negative sign does not come from the cosine, but from the relation $V=-W$. – Steeven Oct 30 '17 at 23:34
• The relation $V=-W$ comes from the fact that $cos(180)=-1$. Look at this – GeeJay Nov 1 '17 at 12:25

The force exerted on a unit positive charge is $\dfrac {kq}{r^2} (+\hat r)$ and so an external force $\dfrac {kq}{r^2} (-\hat r)$ must be applied to move the charge.

If the step is $d\vec r$ then the work done by the external force in moving that step is $\dfrac {kq}{r^2} (-\hat r) \cdot d\vec r = -\dfrac {kq}{r^2} dr$.
You have moved from $\vec r$ to $\vec r + d \vec r$.
Now this must be correct because it says that if $dr$ is positive ie moving away from the positive charge the work done by an external force is negative ie the potential has dropped.
If you move towards the charge then $dr$ is negative and the work done by the external force is positive ie the potential has increased.

The limits of integration will define which way you actually move.

So the potential $\displaystyle V(r) = \int _\infty ^ r-\dfrac {kq}{r^2} dr = + \dfrac {kq}{r}$

Later in answer to @mritunjay 's question.

For many people the sign of $dr$ is conceptually difficult.

Suppose that you were asked to do the following integration $\displaystyle \int _{-1} ^ 0 x\;dx$ which when evaluated is $+\frac 1 2$.

If you look at the left-hand graph below you have found $\frac 1 2 \Delta y \Delta x$ with $\Delta y = y_{\text{final}} - y_{\text{initial}} = -1 -0 = -1$ and $\Delta x = x_{\text{final}} - x_{\text{initial}} = -1 -0 = -1$.

Noting that $x$ is decreasing would you have evaluated the integral $\displaystyle \int _{-1} ^ 0 x\;(-dx)$ instead? The right-hand graph and you are evaluating the work done $F_{\text{external}} \Delta r$.

$F_{\text{external}}$ is negative and with the limits of integration going from $r=\infty$ to $r=r$, $\Delta x$ will also be negative and so you will get a positive result for the work done by the external force.

Note that if the work done by the electric field $E \Delta r$ was evaluated then integration would yield the expected negative answer as $E$ is positive.

• if dr is negative I think BEFORE INTEGRATING u will get V=(kq/(r^2)) dr. But when u integrate this equation with limits u will get V= -(kq/r). If u are sure about ur last step can u please edit ur answer amd elaborate it. Sorry for any inconvenience – Perspicacious Jul 1 '16 at 7:31
• @MritunJay I have expanded my answer to show that the integrations limits take care of sign of $dr$. – Farcher Jul 1 '16 at 12:12
• I believe you are answering for the wrong force in this answer here. Although the OP mentions both the external force and the electric field force, he is trying to calculate the work done by the electric field in the steps he shows. The work done by the electric field is negative. In your answer here, you get a positive work because you find the work done by the external force. That is why you can use $V=W$. But the OP actually asks about the work done by the field, so the relation is opposite: $V=-W$. – Steeven Oct 30 '17 at 23:44