(a) Figure shows two point clusters of charge situated in free space placed on a line that is called the x-axis. The first, with a positive charge of Q1 =+8e, is at the origin......

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Problem:

(a) Figure shows two point clusters of charge situated in free space placed on a line that is called the x-axis. The first, with a positive charge of $Q_1 =+8e$ , is at the origin. The second, with a negative charge of $Q_2 =−4e$ ,is to the right at a distance equal to 0.2m.
(i) What is the magnitude of the force between them?
(ii) Where would you expect to find the position of zero electric field: to the left of $Q_1$ , between $Q_1$ and $Q_2$ or to the right of $Q_2$ ? Briefly explain your choice and then work out the exact position.
(b) The electron in a hydrogen atom orbits the proton at a radius of $5.3\times 10^{-11}$ m.
(i) What is the proton’s electric field strength at the position of the electron?
(ii) What is the magnitude of the electric force on the electron?

Explanation:

For part (a):
(i) The electric force between two charges is calculated using Coulomb’s law:

$F = k_e \dfrac{|Q_1 Q_2|}{r^2}$

where $k_e = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2$ is the Coulomb constant, and $e = 1.6 \times 10^{-19} \, \text{C}$ is the elementary charge.

(ii) The electric field is zero at a point where the field contributions from both charges cancel. This occurs closer to the smaller charge because the electric field’s magnitude decreases with distance.

For part (b):
(i) The electric field strength due to a charge is given by:

$E = k_e \dfrac{|Q|}{r^2}$

where $Q = e = 1.6 \times 10^{-19} \, \text{C}$ .

(ii) The electric force acting on a charge in an electric field is given by:

$F = qE$

where $q = -e = -1.6 \times 10^{-19} \, \text{C}$ .

Solution:

Part (a):

(i) Force between $Q_1$ and $Q_2$ :

$Q_1 = +8e = 8 \times 1.6 \times 10^{-19}$

$= 1.28 \times 10^{-18} \, \text{C}$

$Q_2 = -4e = -4 \times 1.6 \times 10^{-19}$

$= -6.4 \times 10^{-19} \, \text{C}$

$F = k_e \dfrac{|Q_1 Q_2|}{r^2}$

$= (8.99 \times 10^9) \dfrac{(1.28 \times 10^{-18})(6.4 \times 10^{-19})}{(0.2)^2}$

$F = (8.99 \times 10^9) \dfrac{8.192 \times 10^{-37}}{0.04}$

$F = (8.99 \times 10^9)(2.048 \times 10^{-35}) \[= 1.841 \times 10^{-25} \, \text{N}$

Final Answer: $F = 1.841 \times 10^{-25} \, \text{N}$

(ii) Position of zero electric field:
The zero-field point lies between $Q_1$ and $Q_2$ because the charges have opposite signs. Let the distance of the zero-field point from $Q_1$ be $x$ . Then the distance from $Q_2$ is $(0.2 - x)$ .

The electric field at the zero point due to $Q_1$ and $Q_2$ must be equal in magnitude:

$E_1 = E_2 \quad \Rightarrow \quad \dfrac{k_e |Q_1|}{x^2} = \dfrac{k_e |Q_2|}{(0.2 - x)^2}$

$\dfrac{|Q_1|}{x^2} = \dfrac{|Q_2|}{(0.2 - x)^2} \quad \Rightarrow \quad \dfrac{8}{x^2} = \dfrac{4}{(0.2 - x)^2}$

$2 = \dfrac{(0.2 - x)^2}{x^2} \quad \Rightarrow \quad 2x^2 = (0.2 - x)^2$

$2x^2 = 0.04 - 0.4x + x^2 \quad \Rightarrow \quad x^2 + 0.4x - 0.04 = 0$

Solving this quadratic equation:

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-0.4 \pm \sqrt{(0.4)^2 - 4(1)(-0.04)}}{2(1)}$

$x = \dfrac{-0.4 \pm \sqrt{0.16 + 0.16}}{2} = \dfrac{-0.4 \pm 0.5657}{2}$

$x = \dfrac{0.1657}{2} = 0.08285 \, \text{m} \quad \text{(discarding negative root)}$

Final Answer: The zero electric field is at $x = 0.08285 \, \text{m}$ from $Q_1$ .

Part (b):

(i) Electric field strength at the electron’s position:

$E = k_e \dfrac{|Q|}{r^2} = (8.99 \times 10^9) \dfrac{1.6 \times 10^{-19}}{(5.3 \times 10^{-11})^2}$

$E = 5.12 \times 10^{11} \, \text{N/C}$

Final Answer: $E = 5.12 \times 10^{11} \, \text{N/C}$

(ii) Electric force on the electron:

$F = qE = (1.6 \times 10^{-19})(5.12 \times 10^{11}) = 8.19 \times 10^{-8} \, \text{N}$

Final Answer: $F = 8.19 \times 10^{-8} \, \text{N}$

(a) Figure shows two point clusters of charge situated in free space placed on a line that is called the x-axis. The first, with a positive charge of Q1 =+8e, is at the origin……

Problem:

Explanation:

Solution:

Part (a):

Part (b):

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Problem:

Explanation:

Solution:

Part (a):

Part (b):

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