Question

Suppose that the potential difference in going from location $\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{,}\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{,}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{)}$m to locationrole="math" localid="1657094482453" $(2.00,3.52,4.00)$m is$\mathbf{3}\mathbf{V}$. What is the approximate value of${\mathbf{E}}_{\mathbf{y}}$in this region? Include the appropriate sign.

Short answer

The value of the electric field is $-150N/C$.

Step-by-step solution

Electric field and Potential difference

For a charged particle the electric field is described as the ‘negative gradient’ of the electric potential of the particle.

So the sign of the potential difference as well as electric field is important because it indicates whether potential energy of the particle is increasing or decreasing.

Given data

The first location of potential difference is,$\left(2.00,3.50,4.00\right)$.

The last locationpotential differenceis, $\left(2.00,3.52,4.00\right)$.

The potential difference between two locations is, $∆V=3V$.

The value of the electric field

The distance travelled between two locations is given by,

$$\begin{gathered} ∆y=\left(3.52-3.50\right)m \\ ∆y=0.02m \end{gathered}$$

Then, the formula for the electric field $E_y$in is given by,

$$\begin{gathered} ∆V=-E_y\times ∆y \\ {E}_y=\frac{-∆V}{∆y} \\ {E}_y=\frac{-3V}{0.02m}\frac{1N/C}{1V/C} \\ {E}_y=-150N/C \end{gathered}$$

Hence, the value of the electric field is $-150N/C$.