Answer:
Explanation:
4μC will attract -9μC towards the centre and -5μC will repel it away from the centre. Both these forces are opposite to each other.
Force due to 4μC on -9μC towards the centre
= k x Q₁ Q₂/R² = 9 X 10⁹ X 4 X 10⁻⁶ X 9 X 10⁻⁶ / (1.2)² = 225 X 10⁻³ N/C
Force due to -5μC on -9μC away from the centre
= 9 x 10⁹ x 5 x 10⁻⁶x 9 x 10⁻⁶/( 0.8)² = 632.8 x 10⁻³ .N/C
Ner field =407.8 N/C.
Answer:
Explanation:
The computation of the weight of the paper in newtons is shown below:
On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.
Therefore the paper charge is
Now the distance from the charge is
Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.
I’ve answered this before so I know the question is missing an
important given and that given is: <span>1 has an
empty trailer and the other has a fully loaded one.
So, it would be the fully loaded trailer that would take a longer distance to
stop because a lot of weight is being pulled, and when the brakes are started,
the fully loaded trailer is more like pushing against the truck.</span>
Answer: Option (a) is the correct answer.
Explanation:
When these two conducting spheres are connected together through a thin wire then charge from the smaller sphere will travel through the wire. And, this charge will continue to travel towards the neutral sphere until the charge on both the spheres will become equal to each other.
For example, charge on small sphere is 5 C then this charge will continue to travel towards the neutral sphere until its charge also becomes equal to 5 C.
Hence, then their potential will also become equal.
Thus, we can conclude that the spheres are connected by a long, thin wire, then after a sufficiently long time the two spheres are at the same potential.
