Question

(a) In Figure 16.14, what is the direction of the electric field? Is$\mathbf{∆}\mathbf{V}\mathbf{=}{\mathbf{V}}_f\mathbf{-}{\mathbf{V}}_{\mathbf{i}}$positive or negative? (b) In figure 16.15, what is the direction of the electric field? Is$\mathbf{∆}\mathbf{V}\mathbf{=}{\mathbf{V}}_f\mathbf{-}{\mathbf{V}}_{\mathbf{i}}$positive or negative?

Short answer

(a) The value of $∆\mathrm{V}={\mathrm{V}}_f-{\mathrm{V}}_{\mathrm{i}}$is positive and the direction of the electric field is from ‘positive to negative’

(b) The value of electric field and $∆\mathrm{V}={\mathrm{V}}_f-{\mathrm{V}}_{\mathrm{i}}$is zero.

Step-by-step solution

Electric Field Direction

The direction of the electric field for a moving charged particle relies upon the change in the value of the potential difference.

If the electric potential increases from one point to another then the direction of the electric field would be opposite to that of electric potential.

(a) The direction of the electric field

In the figure (a), the term$i$represents the initial position and $f$represents the final position of the particle.

From the given Figure (a), the electric charge moves between two charged plates from the negative region to the positive region.

So, the value of the potential difference between two positions of the particle is given by,

$$\begin{gathered} ∆\mathrm{V}={\mathrm{V}}_f-{\mathrm{V}}_{\mathrm{i}} \\ ∆\mathrm{V}>0\left(\mathrm{positive}\right) \end{gathered}$$

Hence, the value of$∆\mathrm{V}={\mathrm{V}}_f-{\mathrm{V}}_{\mathrm{i}}$is positive.

The value of the potential difference increases from initial to final position, so the direction of the electric field would be opposite to the potential difference.

Hence, the direction of the electric field is from ‘positive to negative’.

(b) The direction of the electric field

From the given Figure (b), the electric charge moves from the negative to the positive inside a closed region.

Inside a closed conductor, the value of the electric field as well as the value of the potential difference is zero.

So, the value of the potential difference between two positions of the particle is given by,

$$\begin{gathered} ∆\mathrm{V}={\mathrm{V}}_f-{\mathrm{V}}_{\mathrm{i}} \\ ∆\mathrm{V}=0 \end{gathered}$$

Hence, the value of electric field and $∆\mathrm{V}={\mathrm{V}}_f-{\mathrm{V}}_{\mathrm{i}}$is zero.