Uniform electric fields
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State the equation that relates electric field strength (E), potential difference (ΔV), and distance (d) in a uniform electric field.
E = ΔV / d, where E is the electric field strength (V/m or N/C), ΔV is the potential difference (V), and d is the distance between the plates (m). This equation is specific to uniform electric fields.
Describe the motion of a positively charged particle placed in a uniform electric field.
A positively charged particle will experience a force in the direction of the electric field. It will accelerate in that direction with a constant acceleration, assuming the electric field is the only force acting on the particle.
A potential difference of 500V is applied across two parallel plates separated by 0.05m. Calculate the electric field strength between the plates.
Using E = ΔV / d, E = 500V / 0.05m = 10,000 V/m. The electric field strength is 10,000 V/m (or 10,000 N/C).
How does the electric field strength change between parallel plates if the separation distance doubles, while the potential difference remains constant?
Since E = ΔV / d, if d doubles and ΔV remains constant, the electric field strength E is halved.
Describe the effect of a uniform electric field on the vertical motion of an electron initially moving horizontally through the field (assume no gravity).
The electron will experience a constant upward force (opposite to the field direction). This results in a constant upward acceleration, causing the electron to follow a parabolic trajectory, similar to projectile motion.
Explain why the electric field between two parallel plates is considered uniform.
The electric field is uniform because the field lines are parallel and equally spaced between the plates (excluding edge effects). This means the electric field strength and direction are the same at all points between the plates.
If a proton is accelerated from rest through a potential difference of 1000V, what is its kinetic energy gain?
The kinetic energy gained by the proton is equal to the work done on it by the electric field, which is qΔV. Therefore, KE gain = (1.60 x 10⁻¹⁹ C)(1000 V) = 1.60 x 10⁻¹⁶ J.
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