Question

In: Physics

Determine the magnitude and direction of the electric field at point 1 in the figure(Figure 1).

Figure 1

Part A

Determine the magnitude and direction of the electric field at point 1 in the figure(Figure 1).

E1→=(2500V/m,up)
E1→=(7500V/m,up)
E1→=(3750V/m,down)
E1→=(2500V/m,down)

Part B

Determine the magnitude and direction of the electric field at point 2 in the figure.

E2→=(2500V/m,up)
E2→=(3750V/m,down)
E2→=(7500V/m,down)
E2→=(5000V/m,up)

Solutions

Expert Solution

Part A

The relation between electric field and potential is, \(E=-\frac{\Delta V}{d}\)

The magnitude of the electric field at point 1 is, \(E_{1}=\left|-\left(\frac{50 \mathrm{~V}-0 \mathrm{~V}}{2 \mathrm{~cm}}\right)\right|\)

\(=\left(\frac{50 \mathrm{~V}-0 \mathrm{~V}}{2 \mathrm{~cm}}\right)\)

\(=\frac{50 \mathrm{~V}}{2 \times 10^{-2} \mathrm{~m}}\)

\(=2500 \mathrm{~V} / \mathrm{m}\)

The direction of the electric field is from high potential to low potential, so the direction of the electric field at point 1 is downwards.

Part B The magnitude of the electric field at point 2 is, \(E_{1}=\left|-\left(\frac{50 \mathrm{~V}-0 \mathrm{~V}}{1 \mathrm{~cm}}\right)\right|\)

\(=\left(\frac{50 \mathrm{~V}-0 \mathrm{~V}}{1 \mathrm{~cm}}\right)\)

\(=\frac{50 \mathrm{~V}}{1 \times 10^{-2} \mathrm{~m}}\)

\(=5000 \mathrm{~V} / \mathrm{m}\)

The direction of the electric field is from high potential to low potential, so the direction of the electric field at point 1 is upwards.

 

Part B

The magnitude of the electric field at point 2 is, 

\(\begin{aligned} E_{1} &=\left|-\left(\frac{50 \mathrm{~V}-0 \mathrm{~V}}{1 \mathrm{~cm}}\right)\right| \\ &=\left(\frac{50 \mathrm{~V}-0 \mathrm{~V}}{1 \mathrm{~cm}}\right) \\ &=\frac{50 \mathrm{~V}}{1 \times 10^{-2} \mathrm{~m}} \\ &=5000 \mathrm{~V} / \mathrm{m} \end{aligned}\)

The direction of the electric field is from high potential to low potential, so the direction of the electric field at point 1 is upwards.

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