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Physics Chapter 10 : Topic 7
Coulomb’s Law
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Coulomb’s law describes the force between:
Reasoning: Coulomb’s law gives the electrostatic force between two point charges.
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According to Coulomb’s law, the electrostatic force between two point charges is directly proportional to:
Reasoning: $\mathbf{F \propto}\mathbf{Q}_{\mathbf{1\ }}\mathbf{Q}_{\mathbf{2}}$ $\mathbf{\Longrightarrow}$ The force increases with the product of the charges.
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According to Coulomb’s law, the electrostatic force is inversely proportional to:
Reasoning: $\mathbf{F \propto}$ $\frac{\mathbf{1}}{\mathbf{r}^{\mathbf{2}}}$
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The mathematical expression for Coulomb’s law is:
Reasoning: Coulomb’s law states: $\mathbf{F = k}$ $\frac{\mathbf{Q}_{\mathbf{1\ }}\mathbf{Q}_{\mathbf{2}}}{\mathbf{r}^{\mathbf{2}}}$
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The value of Coulomb’s constant in free space is approximately:
Reasoning: In free space: $\mathbf{k =}$ $\frac{\mathbf{1}}{\mathbf{4\pi}\mathbf{\ \varepsilon}_{\mathbf{o}}}\mathbf{= 9 \times}\mathbf{10}^{\mathbf{9}}$
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The SI unit of electric charge is:
Reasoning: Electric charge is measured in coulombs $\mathbf{(C)}$.
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The constant $\mathbf{(k)}$ in Coulomb’s law depends on:
Reasoning: $\mathbf{(k)}$ depends on the permittivity of the medium.
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Two charges of the same sign placed close together will:
Reasoning: Like charges repel according to Coulomb’s law.
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Two charges of opposite signs will:
Reasoning: Opposite charges attract.
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If one charge is doubled while the other remains constant, the force becomes:
Reasoning: $\mathbf{F \propto}\mathbf{Q}_{\mathbf{1\ }}\mathbf{Q}_{\mathbf{2}}$ $\mathbf{\Longrightarrow}$ Doubling one charge doubles the force.
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If both charges are doubled, the electrostatic force becomes:
Reasoning: $\left( \mathbf{2}\mathbf{Q}_{\mathbf{1}} \right)\left( \mathbf{2}\mathbf{Q}_{\mathbf{2}} \right)\mathbf{= 4\ }\mathbf{Q}_{\mathbf{1\ }}\mathbf{Q}_{\mathbf{2}}$
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If the distance between two charges is doubled, the force becomes:
Reasoning: $\mathbf{F \propto}$ $\frac{\mathbf{1}}{\mathbf{r}^{\mathbf{2}}}$ $\mathbf{\Longrightarrow}$ Doubling distance reduces force by four times.
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If the separation between two charges is reduced to half, the force becomes:
Reasoning: $\mathbf{F \propto}$ $\frac{\mathbf{1}}{\mathbf{r}^{\mathbf{2}}}$ $\mathbf{\Longrightarrow}$ Halving distance increases force four times.
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Two point charges $\mathbf{Q}_{\mathbf{1}}$ and $\mathbf{Q}_{\mathbf{2}}$ are separated by distance $\mathbf{(r)}$. Theforce depends on:
Reasoning: Coulomb’s law includes both charges and separation distance.
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Two charges $\mathbf{(2C)}$ and $\mathbf{(3C)}$ are separated by $\mathbf{1\ m}$ in free space. The force between them is:
Reasoning: $\mathbf{F =}$ $\frac{\mathbf{k\ }\mathbf{Q}_{\mathbf{1\ }}\mathbf{Q}_{\mathbf{2}}}{\mathbf{r}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{(9\ \times \ }\mathbf{10}^{\mathbf{9}}\mathbf{)(2)(3)}}{\mathbf{1}^{\mathbf{2}}}\mathbf{= 5.4 \times}\mathbf{10}^{\mathbf{10}}\mathbf{\ N}$
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Two charges $\mathbf{1\ \mu C}$ and $\mathbf{2\ \mu C}$ are separated by $\mathbf{1\ m}$. The force between them is:
Reasoning: $\mathbf{F =}\frac{\mathbf{(9\ \times \ }\mathbf{10}^{\mathbf{9}}\mathbf{)(1\ \times \ }\mathbf{10}^{\mathbf{- 6}}\mathbf{)(2\ \times \ }\mathbf{10}^{\mathbf{- 6}}\mathbf{)}}{\mathbf{1}^{\mathbf{2}}}\mathbf{= 18 \times}\mathbf{10}^{\mathbf{- 3}}\mathbf{\ N}$
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Coulomb’s law applies most accurately when charges are:
Reasoning: Coulomb’s law is for point charges.
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The force between two charges in free space is maximum when:
Reasoning: Force increases with charge and decreases with distance.
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The force calculated using Coulomb’s law is:
Reasoning: Like charges repel and unlike charges attract.
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The force between two charges is reduced when they are placed in a medium with:
Reasoning: Higher permittivity reduces electrostatic force.
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In Coulomb’s law, $\mathbf{(r)}$ represents:
Reasoning: $\mathbf{(r)}$ is the separation between the charges.
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Two charges exert forces on each other. According to Newton’s third law:
Reasoning: Charges exert equal magnitude forces in opposite directions.
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If the value of $\mathbf{Q}_{\mathbf{1}}$ becomes zero, the force between charges becomes:
Reasoning: $\mathbf{F = k}$ $\frac{\mathbf{Q}_{\mathbf{1\ }}\mathbf{Q}_{\mathbf{2}}}{\mathbf{r}^{\mathbf{2}}}$ $\mathbf{\Longrightarrow}$ If $\mathbf{Q}_{\mathbf{1}}\mathbf{= 0}$, then $\mathbf{F = 0}$.
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Coulomb’s law is similar to Newton’s law of gravitation because both:
Reasoning: Both forces follow inverse square dependence.
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The direction of electrostatic force between two charges is:
Reasoning: Coulomb force acts along the line joining the two charges.
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Two charges experience a force of $\mathbf{20\ N}$. If their distance becomes three times, the new force will be:
Reasoning: $\mathbf{F \propto}$ $\frac{\mathbf{1}}{\mathbf{r}^{\mathbf{2}}}$ $\mathbf{\Longrightarrow}$ Distance $\mathbf{\times 3}$ gives force divided by $\mathbf{9}$.
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Two point charges have charges $\mathbf{( + \ Q)}$ and $\mathbf{( - \ Q)}$. The force between them is:
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The value of permittivity of free space $\mathbf{(}\mathbf{\varepsilon}_{\mathbf{o}}\mathbf{)}$ affects:
Reasoning: $\mathbf{k =}$ $\frac{\mathbf{1}}{\mathbf{4\pi\ }\mathbf{\varepsilon}_{\mathbf{o}}}$
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If the charge on both particles is increased, the electrostatic force:
Reasoning: Force depends directly on the product of charges.
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Coulomb’s law is valid for charges in free space because:
Reasoning: Free space has permittivity $\mathbf{(0)}$, giving the standard value of $\mathbf{(k)}$.
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