The electric potential (V ) and electric field (⃗ E) are closely related concepts in electrostatics. The electric field is a vector quantity that represents the force per unit charge at a given point in space, whereas electric potential is a scalar quantity that represents the potential energy per unit charge at a given point in space. Electric field and electric potential are related by the equation i.e., electric field is the negative gradient of the electric potential. This means that electric field points in the direction of decreasing potential and its magnitude is the rate of change of potential with distance. The electric field is the force that drives a unit charge to move from higher potential region to lower potential region and electric potential difference between the two points determines the work done in moving a unit charge from one point to the other point. A pair of square conducting plates having sides of length 0.05 m are arranged parallel to each other in the x–y plane. They are 0.01 m apart along the z-axis and are connected to a 200 V power supply as shown in the figure. An electron enters with a speed of 3 × 107 m s−1 horizontally and symmetrically in the space between the two plates. Neglect the effect of gravity on the electron.
Question: 1
The electric field \( \vec{E} \) in the region between the plates is:
Show Hint
Between parallel plates:
\[
E = \frac{V}{d}
\]
Field direction is always from higher potential to lower potential.
Concept:
For parallel plates:
\[
E = \frac{V}{d}
\]
Direction:
Electric field points from higher potential plate to lower potential plate.
Step 1: Calculate magnitude.
\[
V = 200 \, \text{V}, \quad d = 0.01 \, \text{m}
\]
\[
E = \frac{200}{0.01} = 2 \times 10^4 \, \text{V/m}
\]
Step 2: Determine direction.
From the figure, field is along +z direction.
Unit vector along z-axis is \( \hat{k} \).
Final Answer:
\[
\vec{E} = 2 \times 10^4 \, \hat{k} \, \text{V/m}
\]
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Approach Solution -2
The electric field between the plates has to be found from the potential difference applied across them and their separation, then assigned the correct direction as a vector along the given coordinate axis. With \( V = 200\,\text{V} \) and \( d = 0.01\,\text{m} \) between the plates, let's check each option against both the magnitude \( E = V/d \) and the direction implied by which plate is at the higher potential.
\( \left(2\times 10^{2}\,\text{V/m}\right)\hat{k} \): This magnitude would only follow from \( E = V/d \) if the plate separation were mistakenly taken as \( d = 1\,\text{m} \) instead of \( 0.01\,\text{m} \), giving \( 200/1 = 200\,\text{V/m} \) -- two orders of magnitude too small for the actual 1 cm gap. This option is incorrect.
\( -\left(2\times 10^{2}\,\text{V/m}\right)\hat{k} \): This carries the same magnitude error as above and additionally reverses the sign, compounding two mistakes at once. It is incorrect.
\( \left(2\times 10^{4}\,\text{V/m}\right)\hat{k} \): Using the correct separation, \( E = V/d = 200/0.01 = 2\times 10^{4}\,\text{V/m} \). The field inside a parallel-plate arrangement points from the plate at higher potential toward the plate at lower potential; with the higher-potential plate on the \( -z \) side of the gap here, the field points along \( +\hat{k} \). Both the magnitude and the sign are correct.
\( -\left(2\times 10^{4}\,\text{V/m}\right)\hat{k} \): The magnitude here is correct, but the sign is flipped -- this would be the result only if the higher-potential plate were on the \( +z \) side instead, which is not the case in this arrangement. It is incorrect.
Only the option with magnitude \( 2\times 10^{4}\,\text{V/m} \) directed along \( +\hat{k} \) satisfies both the correct plate separation and the correct field direction.
Therefore, the correct answer is \( \left(2\times 10^{4}\,\dfrac{V}{m}\right)\hat{k} \).
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Question: 2
In the region between the plates, the electron moves with an acceleration \( \vec{a} \) given by:
Show Hint
For electrons:
Force opposite to electric field
Always reverse direction after calculating magnitude
Concept:
Force on a charge in an electric field:
\[
\vec{F} = q\vec{E}
\]
Acceleration:
\[
\vec{a} = \frac{q\vec{E}}{m}
\]
For electron:
\[
q = -e
\]
Step 1: Electric field.
From previous result:
\[
\vec{E} = 2 \times 10^4 \, \hat{k} \, \text{V/m}
\]
Step 2: Use electron charge and mass.
\[
e = 1.6 \times 10^{-19} \, \text{C}, \quad
m_e = 9.1 \times 10^{-31} \, \text{kg}
\]
\[
a = \frac{eE}{m}
= \frac{1.6 \times 10^{-19} \times 2 \times 10^4}{9.1 \times 10^{-31}}
\]
Step 3: Calculate magnitude.
\[
a = \frac{3.2 \times 10^{-15}}{9.1 \times 10^{-31}}
\approx 3.5 \times 10^{15} \, \text{m s}^{-2}
\]
Step 4: Direction.
Electron has negative charge, so acceleration is opposite to field.
Field is along \( +\hat{k} \) → acceleration along \( -\hat{k} \).
Final Answer:
\[
\vec{a} = -3.5 \times 10^{15} \, \hat{k} \, \text{m s}^{-2}
\]
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Approach Solution -2
The acceleration of the electron between the plates follows from Newton's second law applied to the electric force it experiences, \( \vec{a} = \dfrac{q\vec{E}}{m} \), using the electron's charge and mass along with the field found for this region, \( \vec{E} = 2\times 10^{4}\,\hat{k}\,\text{V/m} \). Let's check each option against the correct magnitude and direction.
\( -\left(3.5\times 10^{15}\,\text{m s}^{-2}\right)\hat{k} \): The magnitude of the force is \( F = eE = (1.6\times 10^{-19})(2\times 10^{4}) = 3.2\times 10^{-15}\,\text{N} \), and the resulting acceleration is \( a = F/m_e = (3.2\times 10^{-15})/(9.1\times 10^{-31}) \approx 3.5\times 10^{15}\,\text{m s}^{-2} \), matching this option's magnitude. Since the electron's charge is negative, its force (and acceleration) points opposite to \( \vec{E} \), i.e., along \( -\hat{k} \), which also matches. This option is correct.
\( \left(3.5\times 10^{15}\,\text{m s}^{-2}\right)\hat{k} \): The magnitude here is right, but the direction is along \( +\hat{k} \), which would only be correct for a positive charge; for an electron, the acceleration must oppose the field direction. Incorrect.
\( \left(3.5\times 10^{13}\,\text{m s}^{-2}\right)\hat{i} \): This is off by two full orders of magnitude from the correct \( 10^{15} \) value, and it also points along the wrong axis -- the force on the electron acts along the field direction (\( \hat{k} \)), not along \( \hat{i} \), since there is no field component along \( x \) in this region. Incorrect.
\( -\left(3.5\times 10^{13}\,\text{m s}^{-2}\right)\hat{i} \): This shares the same two errors as the previous option -- wrong magnitude by a factor of 100 and the wrong axis -- with the sign flipped as well. Incorrect.
Only the option with magnitude \( 3.5\times 10^{15}\,\text{m s}^{-2} \) directed along \( -\hat{k} \) is consistent with both the size of the electric force and the negative charge of the electron.
Therefore, the correct answer is \( -\left(3.5\times 10^{15}\,\text{m s}^{-2}\right)\hat{k} \).
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Question: 3
Time interval during which an electron moves through the region between the plates is:
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If electric field is perpendicular to motion:
Horizontal velocity remains constant
Time = length / horizontal velocity
Concept:
Electron enters horizontally with velocity along x-axis and experiences:
No force in horizontal direction → uniform motion
Constant vertical acceleration due to electric field
This produces projectile-like motion.
Step 1: Nature of motion.
The motion is similar to:
Uniform velocity in x-direction
Uniform acceleration in vertical direction
Hence, trajectory is a parabola.
Step 2: Direction of deflection.
From earlier results:
Electric field is along \( +\hat{k} \)
Electron (negative charge) accelerates opposite → downward
Step 3: Identify correct path.
The path should:
Start horizontally
Curve downward gradually (parabolic path)
Among the options, only path c shows downward curvature.
Final Answer: Path c
