F = k(Q₁Q₂)/r²
where k is Coulomb's constant 8.99e9
F = 8.99e9(2.8e-6)²/0.5² = 0.2819N
Answer: Magnitude of electrical force stays the same.
Explanation:
Equation:

Since the magnitude of each charge is halved.
and
the separation is halved.


Cancel out .25 on the numerator and denominator. Leaving the original equation.
Answer:
Part a)

Part b)

Explanation:
Part a)
Since the two magnetic field is in same direction
so the net magnetic field is algebraic sum of magnetic field due to both
so here magnetic field of wire is given as

here we know that
I = 2 A
r = 5 cm
so we will have


So net magnetic field is given as

Part b)
When direction of current is reversed then the direction of magnetic field is also reversed
So we will have

Answer:
Explanation:
A. Using
E= ma/q
E=m/q(2s/t²)
So
E= 9.1x10^-31/1.6*10^-19( 2*4.5/ 3*10-12)
E=5.7NC
The electric field has to be downward since the force is positive that is upward
B.
The electron acceleration is of the order of 10^11 times greater so for practical purposes we neglect the effect of gravity
W^2 = 2 A 8 where A is the angular acceleration.
(3w)^2 = 2 A R where R is the number of revolutions.
Note that you are asked for the additional revolutions.