Question

1. The electron and proton atom are separated ( on the average) by a distance of 5.3 × 10-11m. find the magnitude of the electric force and the gravitational force between the two particles.
2. Calculate the potential difference between two points x,y in the field of a single charge of
1.5 × 10-6C if the products and the differences of the distance of x and y from the 15cm and 2cm respectively
3. A charge 4.5 × 10-9C is placed in an electric field of magnitude 9.6 × 10-5 NC-1 upwardly directed. What work is done by the electric force in moving the charge 60cm to the left.

4. A charge 4µC is place 60cm away from a charge of 4µC what is the electric field at point?
a. p midway between the charges
b. at point q 4cm from p and equidistance from the two charges

Answer 1

Answer on Question #79342, Physics / Electric Circuits

1. The electron and proton atom are separated ( on the average) by a distance of $5.3 \times 10^{-11} \mathrm{~m}$. Find the magnitude of the electric force and the gravitational force between the two particles.

Solution

Magnitude of the electric force between the two particles:

$$F_{el} = \frac{k e^{2}}{r^{2}} = (9 \cdot 10^{9}) \left(\frac{1.6 \cdot 10^{-19}}{5.3 \cdot 10^{-11}}\right)^{2} = 8.2 \cdot 10^{-8} \mathrm{~N}.$$

Magnitude of the gravitational force between the two particles:

$$F_{el} = \frac{G m M}{r^{2}} = (6.67 \cdot 10^{-11}) \frac{(9.1 \cdot 10^{-31}) (1.67 \cdot 10^{-27})}{(5.3 \cdot 10^{-11})^{2}} = 3.6 \cdot 10^{-47} \mathrm{~N}.$$

2. Calculate the potential difference between two points $x, y$ in the field of a single charge of $1.5 \times 10^{-6} \mathrm{C}$ if the products and the differences of the distance of $x$ and $y$ from the 15cm and 2cm respectively

Solution

$$V_{1} - V_{2} = (9 \cdot 10^{9}) (1.5 \cdot 10^{-6}) \left(\frac{1}{0.02} - \frac{1}{0.15}\right) = 585 \mathrm{~kV}.$$

3. A charge $4.5 \times 10^{-9} \mathrm{C}$ is placed in an electric field of magnitude $9.6 \times 10^{-5} \mathrm{NC}^{-1}$ upwardly directed. What work is done by the electric force in moving the charge $60 \mathrm{~cm}$ to the left.

Solution

The work is done by the electric force in moving the charge $60 \mathrm{~cm}$ to the left is zero:

$$W = q E d \cos 90 = q E d(0) = 0 J.$$

4. A charge $4 \mu \mathrm{C}$ is placed 60cm away from a charge of $4 \mu \mathrm{C}$ what is the electric field at point?

a. p midway between the charges

b. at point q 4cm from p and equidistance from the two charges

Solution

a. The electric field at p midway between the equal charges is zero:

$$E = 0 \frac{V}{m}.$$

b.

$$E = 2 \frac {k q}{r ^ {2}} \sin \alpha$$ $$E = \frac {2 (9 \cdot 1 0 ^ {9}) (4 \cdot 1 0 ^ {- 6})}{0 . 0 3 ^ {2} + 0 . 0 4 ^ {2}} \frac {0 . 0 4}{\sqrt {0 . 0 3 ^ {2} + 0 . 0 4 ^ {2}}} = 2 3 \cdot 1 0 ^ {6} \frac {V}{m}.$$

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