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ioda
2 years ago
11

Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way

between the plates?
Physics
1 answer:
VladimirAG [237]2 years ago
3 0

Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that,  F=q*E

The electric field between  the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

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An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i
Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle=\alpha=40.8^{\circ}

Initial speed =v_0=346m/s

We have to find the horizontal distance covered  by the shell after 5.03 s.

Horizontal component of initial speed=v_{ox}=v_0cos\theta=346cos40.8=261.9m/s

Vertical component of initial speed=v_{oy}=346sin40.8=226.1m/s

Time=t=5.03 s

Horizontal distance =Horizontal\;velocity\times time

Using the  formula

Horizontal distance=261.9\times 5.03

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0
3 years ago
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
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Answer:

F=1.26*10^{-3}N

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

F=\frac{kq_1q_2}{d^2}

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(q_1,q_2) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

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3 years ago
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