Figure 16.59 shows several locations inside a capacitor. You need to calculate the potential difference${\mathbf{V}}_{\mathbf{D}}\mathbf{-}{\mathbf{V}}_{\mathbf{C}}$

(a) What is the direction of the path (+x or −x)? (b)If the charge on the right plate is negative and the charge on the left plate is positive, what is the sign of ${\mathbf{V}}_{\mathbf{D}}\mathbf{-}{\mathbf{V}}_{\mathbf{C}}$?

The given data can be listed below,

• The charge on the right plate is, -q
• The charge on the left plate is, -q

The potential at a position is the amount that, when multiplied by the charge of another particle, yields the potential energy of that particle at that location.

The potential difference$V_D-V_C$ is calculated by travelling from point C to D so the direction of path is from -x-direction to +x-direction.

Thus, the direction of the path is +x-direction.

The point C have more potential because of the positive charge on the left plate and the potential on the point D is less because of the negative charge carrying plate. So, the potential difference is negative which is given by,

$∆V=V_D-V_C$

Thus, the sign of the potential difference is negative