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
Answer:
I think it is 5.6. This is my answer
Answer:
induced current
Explanation:
intentionally manipulated.
Answer:
No work is performed or required in moving the positive charge from point A to point B.
Explanation:
Lets take
Q= Positive charge which move from point A to point B along
Voltage difference,ΔV =V₁ - V₂
The work done
W = Q . ΔV
Given that charge is moved from point A to point B along an equipotential surface.It means that voltage difference is zero.
ΔV = 0
So
W = Q . ΔV
W = Q x 0
W= 0 J
So work is zero.
Answer:
3.33 N
Explanation:
First, find the acceleration.
Given:
Δx = 3 m
v₀ = 0 m/s
t = 3 s
Find: a
Δx = v₀ t + ½ at²
3 m = (0 m/s) (3 s) + ½ a (3 s)²
a = ⅔ m/s²
Use Newton's second law to find the force.
F = ma
F = (5 kg) (⅔ m/s²)
F ≈ 3.33 N