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Physics Chapter 10 : Topic 2
Electric Field Strength
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Electric field strength at a point is defined as:
Reasoning: Electric field strength is the force experienced per unit positive test charge. $\mathbf{\Longrightarrow}$ $\mathbf{E =}$ $\frac{\mathbf{F}}{\mathbf{q}}$
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The SI unit of electric field strength is:
Reasoning: From $\mathbf{E =}$ $\frac{\mathbf{F}}{\mathbf{q}}$, the unit is Newton per Coulomb $\mathbf{(N\ /\ C)}$.
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A charge of $\mathbf{2\ C}$ experiences a force of $\mathbf{10\ N}$ in an electric field. The field strength is:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{F}}{\mathbf{q}}\mathbf{=}\frac{\mathbf{10}}{\mathbf{2}}\mathbf{= 5\ N/C}$
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The force acting on a charge in an electric field is given by:
Reasoning: Rearranging $\mathbf{E =}$ $\frac{\mathbf{F}}{\mathbf{q}}$ gives $\mathbf{F = qE}$.
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If the electric field strength at a point is $\mathbf{50\ N/C}$, it means:
Reasoning: Electric field strength represents force per unit charge.
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A charge of $\mathbf{4 \times}\mathbf{10}^{\mathbf{- 6}}\mathbf{\ C}$ is placed in an electric field of $\mathbf{2000\ N/C}$. The force on the charge is:
Reasoning: $\mathbf{F = qE = (4 \times}\mathbf{10}^{\mathbf{- 6}}\mathbf{)(2000) = 8 \times}\mathbf{10}^{\mathbf{- 3}}\mathbf{\ N}$
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If the charge placed in an electric field is doubled, the force on it becomes:
Reasoning: From $\mathbf{(F = qE)}$, force is directly proportional to charge.
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If electric field strength is increased while charge remains constant, the force will:
Reasoning: Since $\mathbf{(F = qE)}$, increasing $\mathbf{(E)}$ increases force.
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A negative charge placed in an electric field experiences force:
Reasoning: Negative charges experience force opposite to the electric field direction.
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Electric field strength is a vector quantity because it has:
Reasoning: Electric field has both magnitude and direction.
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A uniform electric field exists between parallel plates when:
Reasoning: A uniform field has the same strength and direction throughout the region.
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The electric field between two charged parallel plates can be calculated using:
Reasoning: For a uniform electric field: $\mathbf{E =}$ $\frac{\mathbf{\Delta V}}{\mathbf{\Delta d}}$
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Two parallel plates have a potential difference of $\mathbf{100\ V}$ and are separated by$\mathbf{\ }\mathbf{0.5\ m}$. The electric field strength is:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{\Delta V}}{\mathbf{\Delta d}}$ $\mathbf{\Longrightarrow}$ $\mathbf{E =}$ $\frac{\mathbf{100}}{\mathbf{0.5}}\mathbf{= 200\ V/m}$
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The unit volt per metre $\mathbf{(V/m)}$ is equivalent to:
Reasoning: $\mathbf{1\ V/m = 1\ N/C}$
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Two charged plates have a separation of $\mathbf{2\ cm}$ and a potential difference of $\mathbf{400\ V}$. The electric field strength is:
Reasoning: $\mathbf{2\ cm = 0.02\ m}$ $\mathbf{\Longrightarrow}$ $\mathbf{E =}$ $\frac{\mathbf{400}}{\mathbf{0.02}}\mathbf{= 20000\ V/m}$
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In a uniform electric field between parallel plates, the field strength depends on:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{\Delta V}}{\mathbf{\Delta d}}$ depends on voltage difference and distance.
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A test charge is used to measure electric field strength because:
Reasoning: A test charge should be small enough not to disturb the existing field.
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If the distance between parallel plates is increased while voltage remains constant, the electric field strength:
Reasoning: From: $\mathbf{E =}$ $\frac{\mathbf{\Delta V}}{\mathbf{\Delta d}}$ increasing distance decreases field strength.
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If the potential difference between parallel plates is doubled while separation remains constant, the electric field:
Reasoning: Electric field strength is directly proportional to potential difference.
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A particle with charge $\mathbf{5C}$ experiences a force of $\mathbf{25\ N}$. The electric field strength is:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{F}}{\mathbf{q}}\mathbf{=}\frac{\mathbf{25}}{\mathbf{5}}\mathbf{= 5\ N/C}$
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Which change will NOT affect the electric field strength between parallel plates?
Reasoning: Electric field depends on plates, not on the test charge.
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A charge of $\mathbf{3 \times}\mathbf{10}^{\mathbf{- 6}}\mathbf{\ C}$ experiences a force of $\mathbf{6 \times}\mathbf{10}^{\mathbf{- 3}}\mathbf{\ N}$. The electric field strength is:
Reasoning: $\mathbf{E =}$ $\frac{\mathbf{F}}{\mathbf{q}}\mathbf{=}\frac{\mathbf{6\ \times \ }\mathbf{10}^{\mathbf{- 3}}}{\mathbf{3\ \times \ }\mathbf{10}^{\mathbf{- 6}}}$ $\mathbf{= 2000\ N/C}$
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In the equation $\mathbf{E =}$ $\frac{\mathbf{\Delta V}}{\mathbf{\Delta d}}\mathbf{\ }$, $\mathbf{\Delta d}$ represents:
Reasoning: $\mathbf{\Delta d}$ represents the separation over which the potential changes.
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A uniform electric field between plates is directed from:
Reasoning: Electric field lines move from positive to negative plates.
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If a charge is placed in a region where electric field strength is zero, the force on it will be:
Reasoning: $\mathbf{F = qE}$ $\mathbf{\Longrightarrow}$ If $\mathbf{E = 0}$, then $\mathbf{F = 0}$.
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A parallel plate capacitor has voltage $\mathbf{600\ V}$ across plates separated by $\mathbf{3\ cm}$. The electric field is:
Reasoning: $\mathbf{3\ cm = 0.03\ m}$ $\mathbf{\Longrightarrow}$ $\mathbf{E =}$ $\frac{\mathbf{600}}{\mathbf{0.03}}\mathbf{= 20000\ V/m}$
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Which statement correctly relates electric field strength and force?
Reasoning: $\mathbf{F = qE}$ shows force depends on charge and field strength.
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Which factor determines the magnitude of force on a charged particle in an electric field?
Reasoning: Force is calculated by $\mathbf{(F = qE)}$.
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A charge of $\mathbf{(1\ C)}$ placed in an electric field of $\mathbf{(1\ N/C)}$ experiences a force of:
Reasoning: $\mathbf{F = qE = (1)(1) = 1\ N}$
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The electric field strength between parallel plates becomes larger when:
Reasoning: Increasing potential difference increases $\mathbf{=}$ $\frac{\mathbf{\Delta V}}{\mathbf{\Delta d}}$
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