# Electric Field stays the Same even though Distance Increases?

I was working on physics homework and came across this multiple choice question.

Two metal plates of area A and separated by a distance d are placed in parallel near each other to form a capacitor with capacitance C. The plates are connected to a voltage source with potential V and allowed to charge completely. The voltage is then removed, and the plates moved so that they are now separated by a distance 2d.Which of the following statements is true?

Now, one of the answers that was incorrect was electric field decreases, which left me puzzled. From my understanding, charge would stay the same, capacitance would decrease, and therefore voltage would increase. In one way, it makes sense that the electric field of the capacitance would stay the same since voltage increases at the same rate distance decreases. But the other way is that if charge of the capacitors are staying the same, then how is electric field not changing when distance increases? I guess I just gave myself the explanation for my question, but it just isn't clicking. Anyone have a different/better explanation?

First, we have

$$C=\frac{Q}{V}$$

As you said, increasing $$d$$ decreases $$C$$, and since $$Q$$ is unchanged this means $$V$$ increases.

Second we have, for a parallel plate capacitor

$$E=\frac{V}{d}$$

Since both $$V$$ and $$d$$ increase, the electric field $$E$$ remains the same.

Hope this helps.

• Consider the distance D is infinity, clearly there is a decrease in E. This equation is an approximation. so the E field does decrease surely? Commented Nov 27, 2021 at 2:10

Most textbook assume that plates are infinitely long so for infinitely long plates electric field is p/2e where p is charge density on plate and e is permittivity as you can see p is constant electric field independent of distance too remains constant.