5 Marks Questions

Question 15 Marks

Two spherical bodies, each of mass 50kg, are placed at a separation of 20cm. Equal charges are placed on the bodies and it is found that the force of Coulomb repulsion equals the gravitational attraction in magnitude. Find the magnitude of the charge placed on either body.

Answer

Mass = 50kg

r = 20 cm = 0.2m

$\text{F}_\text{G}=\text{G}\frac{\text{m}_1\text{m}_2}{\text{r}^2}$

$=\frac{6.67\times10^{-11}\times2500}{0.04}$

Coulomb's force $\text{F}_\text{c}=\frac{1}{4\pi\varepsilon_\text{o}}\frac{\text{q}_1\text{q}_2}{\text{r}^2}$

$=9\times 10^9\frac{\text{q}^2}{0.04}$

Since. $\text{F}_\text{G}=\text{F}_\text{c}=\frac{6.7\times10^{-11}\times 2500}{0.04}=\frac{9\times10^9\times \text{q}^2}{0.04}$

$\Rightarrow \text{q}^2=\frac{6.7\times10^{-11}\times 2500}{0.04}$

$=\frac{6.7\times 10^{-9}}{9\times 10^{9}}\times25$

$=18.07\times 10^{-18}$

$\text{q}=\sqrt{18.07\times 10^{-18}}$

$4.3\times 10^{-9}\text{C}$

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Question 25 Marks

A satellite is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station? (Radius of the earth is 6400km.)

Answer

Let h be the height, M be the Earth's mass, R be the Earth's radius and m be the satellite's mass

Force on the satellite due to the earth when it is at the Earth's surface, $\text{F}_1=\frac{\text{GMm}}{(\text{R + h})^2}$

Force on the satellite due to the earth when it is at height h above the Earth's surface, $\text{F}_2=\frac{\text{GMm}}{(\text{R + h})^2}$

According to question, we have:

$\frac{\text{F}_1}{\text{F}_2}=\frac{(\text{R + h})^2}{\text{R}^2}$

$\Rightarrow2=\frac{(\text{R + h})^2}{\text{R}^2}$

Taking squareroot on both sides, we get:

$\sqrt{2}=1+\frac{\text{h}}{\text{R}}$

$\Rightarrow\text{h}=(\sqrt{2}-1)\text{R}$

$=0.414\times6400=2649.6\text{km}\approx2650\text{km}$

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Question 35 Marks

Two charged particles placed at a separation of 20cm exert 20N of Coulomb force on each other. What will be the force if the separation is increased to 25cm?

Answer

Two charged particles placed at a separation of 20cm exert 20N of Coulomb force on each other.

So, $\text{F}_1=\frac{1}{4\pi\in_0}\cdot\frac{\text{q}^2}{\text{r}^2_1}$

Also, $\text{F}_2=\frac{1}{4\pi\in_0}\cdot\frac{\text{q}^2}{\text{r}^2_2}$

According to the question, we have:

$\frac{\text{F}_1}{\text{F}_2}=\frac{\text{r}^2_1}{\text{r}^2_2}$

$=\frac{20\times20}{25\times25}=\frac{16}{25}$

$\therefore\text{F}_2=\frac{16}{25}\times\text{F}_1$

$\Rightarrow\text{F}_2=\frac{16}{25}\times20$

$\Rightarrow\text{F}_2=12.8\text{N}\approx13.0\text{N}$

Therefore, the two charged particles will exert a force of 13.0N on each other, if the separation is increased to 25cm.

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