In: Physics
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) |
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.