Question

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 ${\mathbf{V}}_{\mathbf{D}}\mathbf{-}{\mathbf{V}}_{\mathbf{C}}$?

Short answer

a)The direction of the path is +x-direction.

b) The sign of the potential difference is negative.

Step-by-step solution

Identification of given data

The given data can be listed below,

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

Concept/Significance of electric potential.

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.

(a) Determination of the direction of the path

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.

(b) Determination of the sign of VD-VC.

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