9702 · 18.2
Uniform electric fields flashcards
Revision flashcards for Cambridge 9702 Uniform electric fields (syllabus 18.2). Flip, recall, then mark a real past-paper question.
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What defines a uniform electric field?
A region where the electric field strength (E) is constant in both magnitude and direction.
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How are uniform electric fields visually represented?
By parallel, equally spaced electric field lines.
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What is the fundamental definition of electric field strength (E)?
The force per unit positive test charge ($E = F/Q$).
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What is the formula for electric field strength (E) between parallel plates?
$E = \frac{\Delta V}{d}$, where $\Delta V$ is potential difference and $d$ is plate separation.
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What are equipotential surfaces in a uniform electric field?
Lines (or surfaces) parallel to the plates and perpendicular to the electric field lines, representing points of equal electric potential.
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How much work is done by the electric field when a charge moves along an equipotential line?
No work is done.
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In which direction does a negative charge accelerate in a uniform electric field?
Opposite to the direction of the electric field lines.
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Describe the path of a charged particle fired perpendicular to a uniform electric field.
It will follow a parabolic trajectory.
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How is work done (W) related to charge (Q) and potential difference (ΔV) in an electric field?
$W = Q\Delta V$.
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What does the uniform density of electric field lines indicate?
A constant electric field strength throughout the region.
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What is the relationship between work done (W) on a charge and its change in kinetic energy (ΔKE) in an electric field?
The work done by the electric field on the charge is equal to the change in the charge's kinetic energy, so $W = \Delta K_E$. This can be expressed as $q\Delta V = \frac{1}{2}m(v_f^2 - v_i^2)$.
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How do you calculate the acceleration 'a' of a particle with mass 'm' and charge 'q' in a uniform electric field 'E'?
Using Newton's second law, the acceleration is the force divided by mass: $a = \frac{F}{m} = \frac{Eq}{m}$.
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Why is the horizontal velocity of a particle fired perpendicular to a uniform E-field constant?
The electric force acts perpendicular to the initial velocity (i.e., vertically). Since there is no horizontal force, there is no horizontal acceleration, and the horizontal velocity remains constant.
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What are the two equivalent units for electric field strength, E?
Newtons per Coulomb (N C⁻¹) and Volts per metre (V m⁻¹).