The dotted lines represent equipotential and the solid lines represent electric field. potential at a = -50V potential at b = -20V The question is: What is the electric potential at c, the middle point between a and b? And the answer provided is greater than -35V.
Can someone please explain to me why this is so, qualitatively and quantitatively?
$$E=k\frac{q}{r^2}$$
$r$ is the distance to the source of the field $E$, and that source has charge $q$. $k$ is a constant.
And here you see the explanation behind your confusion: An electric field doesn't decrease linearly with distance as you thought, but quadratically. Because of the squared $r^2$.
So for example you can't expect the field to half when the distance is halved. Rather, the field is one fourth when the distance is halved. In your case, we are not halving the distance, but this shows that the logic behind picking the middle-value doesn't hold for quadratic relationships.
The electric field lines are getting closer together towards the right and this tells you that the magnitude of the electric field is larger on the right than on the left.
That being so and the electric filed $E$ being related to the potential gradient $\frac{dV}{dr} = - E$ means the magnitude of the potential gradient is not constant as one move from left to right.
To get $-35$ volts you assume that the potential in the middle is the average of $-2-$ and $-35$ volts ie assume a constant potential gradient.
If the electric field is due to a point charge then the potential will change in proportion to the reciprocal of the distance from the point charge.
So imaging that you are going down a hill starting at $b$ at a height of $20$ metres below the summit with the downward slope getting greater and greater as you reach $a$ which is $50$ metres below the summit.
How far hove you dropped when you reached $c$?
It might help you to draw a sketch graph of height (potential) against position showing this increased downwards steepness going from $b$ to $a$?
On the sketch graph for comparison draw a straight line which would represent a constant potential gradient.
If you assume that $V \propto \frac 1 r$ where $r$ is the distance from the point charge then you can get a numerical value to your problem.
Start with $-50 = \frac {K}{r_1}$ where $K$ is a constant and $r_1$ is the distance of $a$ from the point charge.
