The potential difference between the accelerating plates of a TV set is about 25 kV. If the distance between the plates is 1.5 cm, find the magnitude of the uniform electric field in the region between the plates.
Question
The potential difference between the accelerating plates of a TV set is about 25 kV. If the distance between the plates is 1.5 cm, find the magnitude of the uniform electric field in the region between the plates.
Solution
The magnitude of a uniform electric field (E) can be calculated using the formula:
E = V/d
where: V = potential difference (in volts) d = distance (in meters)
Given in the problem, V = 25 kV = 25,000 volts (since 1 kV = 1000 volts) d = 1.5 cm = 0.015 meters (since 1 cm = 0.01 meters)
Substituting these values into the formula, we get:
E = 25,000 volts / 0.015 meters = 1,666,666.67 volts per meter (V/m)
So, the magnitude of the uniform electric field in the region between the plates is approximately 1.67 x 10^6 V/m.
Similar Questions
The figure below shows an electron entering a uniform electric field perpendicularly betweenthe middle of two plates with a velocity of 5.0 × 107 m s−1. Two flat parallel plates, each oflength 12.0 cm, are separated by a distance of 1.5 cm.The potential difference between the plates is 210 V.(a) Calculate the magnitude of the electric field strength between the plates.(b) Determine the magnitude and direction of the acceleration of the electron.(c) Calculate the time for the electron to travel a horizontal distance equal to the length of theplates.(d) Use your answers in (c) to determine whether the electron will hit one of the plates oremerge from between the plates.
A parallel-plate capacitor is formed of of two 20.0 cm × 20.0 cm plates spaced 2.50 cm apart. The plates are charged to ±±4.00 nC. What is the magnitude of the electric field between the plates?
n a spherical capacitor with plate radii a & b, the potential difference between the plates is V. The electric field between the plates at a radial distance r from centre isChoose answer:
At what rate does the electric field change between the plates of a squarecapacitor of side 5 cm, if the plates are spaced 1.2 mm apart and the voltageacross them is changing at a rate of 60 V/s?
Electrons emitted from the cathode of a cathode ray tube are accelerated by a potentialdifference of 2.00 kV between the cathode and the anode. The electrons then enter the uniformelectric field between the middle of two parallel conducting plates, each of length 2.00 cm. ThePotential difference between the plates is 500 V and the plates are separated by a distance of1.50 cm.Calculate(a) the velocity 𝑢 of the electrons when they emerge from the anode.(b) the time taken for the electrons to transverse the uniform electric field.(c) the velocity of the electrons when they emerge from the uniform electric field.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.