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Physics Chapter 10 : Topic 8
Electric Field Due to a Point Charge
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Electric field strength at a point is defined as:
Reasoning: Electric field strength is the force acting on a unit positive test charge placed at that point.
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The formula for electric field strength due to a point charge in free space is:
Reasoning: The electric field due to a point charge follows the inverse square law. $\mathbf{\Longrightarrow}$ $\mathbf{E =}$ $\frac{\mathbf{kQ}}{\mathbf{r}^{\mathbf{2}}}$
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In the equation $\mathbf{k}\mathbf{=}$ $\frac{\mathbf{kQ}}{\mathbf{r}^{\mathbf{2}}}$, $\mathbf{Q}$ represents:
Reasoning: $\mathbf{Q}$ is the charge creating the electric field.
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The SI unit of electric field strength is:
Reasoning: Electric field strength is force per unit charge.$\mathbf{E =}$ $\frac{\mathbf{F}}{\mathbf{q}}$ ; Unit: $\mathbf{N}\mathbf{/C}$
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Electric field strength at a point depends on:
Reasoning: The field depends on the charge producing it and the distance from that charge.
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If the magnitude of a point charge is doubled, the electric field strength at the same distance becomes:
Reasoning: $\mathbf{E \propto Q}$ $\mathbf{\Longrightarrow}$ Doubling charge doubles field strength.
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If the distance from a point charge is doubled, the electric field strength becomes:
Reasoning: $\mathbf{E \propto}$ $\frac{\mathbf{1}}{\mathbf{r}^{\mathbf{2}}}$ $\mathbf{\Longrightarrow}$ Doubling distance reduces field by four times.
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If the distance from a point charge is reduced to half, the electric field strength becomes:
Reasoning: $\mathbf{E \propto}$ $\frac{\mathbf{1}}{\mathbf{r}^{\mathbf{2}}}$ $\mathbf{\Longrightarrow}$ Halving distance increases field by four times.
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The direction of electric field due to a positive point charge is:
Reasoning: Positive charges repel positive test charges, so the field points outward.
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The direction of electric field due to a negative point charge is:
Reasoning: Negative charges attract positive test charges.
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A point charge produces an electric field that decreases with distance because:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{kQ}}{\mathbf{r}^{\mathbf{2}}}$ $\mathbf{\Longrightarrow}$ Field decreases as distance increases.
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A charge of $\mathbf{(2\ C)}$ produces an electric field at a distance of $\mathbf{1\ m}$. The field strength is:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{kQ}}{\mathbf{r}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{(9\ \times \ }\mathbf{10}^{\mathbf{9}}\mathbf{)(2)}}{\mathbf{1}^{\mathbf{2}}}\mathbf{= 18 \times}\mathbf{10}^{\mathbf{9}}\mathbf{\ N/C}$
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A charge of $\mathbf{5\mu C}$ produces an electric field at $\mathbf{2\ m}$ distance. The field strength is:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{(9\ \times \ }\mathbf{10}^{\mathbf{9}}\mathbf{)(5\ \times \ }\mathbf{10}^{\mathbf{- 6}}\mathbf{)}}{\mathbf{2}^{\mathbf{2}}}\mathbf{= 1.125 \times}\mathbf{10}^{\mathbf{4}}\mathbf{\ N/C}$
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The electric field strength is maximum:
Reasoning: Field strength increases as distance decreases.
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A test charge placed in an electric field experiences a force because:
Reasoning: Electric fields exert forces on charged particles.
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If the test charge used to measure an electric field is doubled, the electric field strength:
Reasoning: Electric field is a property of the source charge and location, not the test charge.
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The value of $\mathbf{(k)}$ in free space is:
Reasoning: Coulomb constant in vacuum is approximately $\mathbf{9 \times}\mathbf{10}^{\mathbf{9\ }}\mathbf{Nm}^{\mathbf{2}}\mathbf{/\ }\mathbf{C}^{\mathbf{2}}$.
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In the formula $\mathbf{E =}$ $\frac{\mathbf{kQ}}{\mathbf{r}^{\mathbf{2}}}$, increasing ‘$\mathbf{r}$’ affects the electric field:
Reasoning: Distance appears squared in the denominator.
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Two points are located at distances $\mathbf{(r)}$ and $\mathbf{(2r)}$ from the same charge. The ratio of electric fields is:
Reasoning: $\mathbf{E}_{\mathbf{1}}\mathbf{=}$ $\frac{\mathbf{kQ}}{\mathbf{r}^{\mathbf{2}}}$ $\mathbf{\Longrightarrow}$ $\mathbf{E}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{kQ}}{\mathbf{(2r)}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{E}_{\mathbf{1}}}{\mathbf{4}}$ $\mathbf{\Longrightarrow}$ So, $\mathbf{E}_{\mathbf{1}}\mathbf{\ :}\mathbf{E}_{\mathbf{2}}\mathbf{= 4\ :1}$
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Electric field strength is zero at a point where:
Reasoning: Zero electric field means no electric force on a test charge.
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A positive charge placed in the electric field of another positive charge moves:
Reasoning: Like charges repel.
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A negative charge placed near a positive point charge moves:
Reasoning: Opposite charges attract.
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The electric field strength due to a point charge becomes four times when:
Reasoning: $\mathbf{E \propto}$ $\frac{\mathbf{1}}{\mathbf{r}^{\mathbf{2}}}$ $\mathbf{\Longrightarrow}$ Halving distance gives four times field.
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If a charge producing the electric field is reduced to half, the field strength:
Reasoning: $\mathbf{E \propto Q}$
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The electric field strength due to a point charge is independent of:
Reasoning: Electric field does not depend on the test charge.
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A point charge produces an electric field of $\mathbf{(100\ N/C)}$ at a distance $\mathbf{(r)}$. At distance $\mathbf{(2r)}$, the field is:
Reasoning: $\mathbf{E \propto}$ $\frac{\mathbf{1}}{\mathbf{r}^{\mathbf{2}}}$ $\mathbf{\Longrightarrow}$ Distance doubled $\mathbf{\Longrightarrow}$ Field becomes one-fourth.
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The electric field at a point due to multiple charges is found by:
Reasoning: Electric fields follow the principle of superposition.
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The electric field around a point charge is represented by field lines that are:
Reasoning: Field lines radiate outward from positive charges and inward toward negative charges.
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The formula $\left( \mathbf{E =}\frac{\mathbf{kQ}}{\mathbf{r}^{\mathbf{2}}} \right)$ is valid for:
Reasoning: Coulomb’s law and electric field formula apply to point charges in free space.
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