Answer:
50 V
Explanation:
The formula electric potential is given as,
V = kq/r............. Equation 1
q = Vr/k ................. Equation 2
Where q = charge at that point, V = Electric potential, k = coulombs constant, r = distant.
Given: V = 100 V, r = 2.0 m, k = 9.0×10⁹ Nm/C².
Substitute into equation 1
q = (100×2)/(9.0×10⁹ )
q = (200/9)(10⁹)
q = 22.22×10⁻⁹
q = 2.22×10⁻⁸ C.
The potential at point 4.0 m
Given: r = 4.0 m, q = 2.22×10⁻⁸ C, k = 9.0×10⁹ Nm²/C²
Substitute into equation 2
V = 9.0×10⁹(2.22×10⁻⁸)/4
V = 49.95 V
V ≈ 50 V
Hence the potential = 50 V
Answer:
Wellll. I am assuming the direction of speed is in the same direction as the direction of displacement of the train. (i.e. Velocity is positive)
Acceleration is defined as the rate of change of velocity with respect to time (m^s-2)
Explanation:
Answer: q2 = -0.05286
Explanation:
Given that
Charge q1 = - 0.00325C
Electric force F = 48900N
The electric field strength experienced by the charge will be force per unit charge. That is
E = F/q
Substitute F and q into the formula
E = 48900/0.00325
E = 15046153.85 N/C
The value of the repelled second charge will be achieved by using the formula
E = kq/d^2
Where the value of constant
k = 8.99×10^9Nm^2/C^2
d = 5.62m
Substitutes E, d and k into the formula
15046153.85 = 8.99×10^9q/5.62^2
15046153.85 = 284634186.5q
Make q the subject of formula
q2 = 15046153.85/ 28463416.5
q2 = 0.05286
Since they repelled each other, q2 will be negative. Therefore,
q2 = -0.05286
The correct answer is true
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