Question

A. Draw the electric field lines around a negative charge.

Answer 1

a. Answer:

We use Electric field lines for visualizing electric  fields, so this helps us to see the problem more real. So an electric field line is an imaginary  line or curve drawn through a region of space such that the tangent at any point comes from the direction of the electric-field vector at that point. The electric field lines around a negative charge is shown in the First figure below.

b. Answer:

Electric forces can be found by using the Coulomb Law's that states that The magnitude of the electric force between two point charges is directly proportional  to the product of the charges and inversely proportional to the square  of the distance between them. This can be expressed as follows:

$F=k\frac{\left | q_{1}q_{2} \right |}{r^2} \\ \\ Where: \\ \\ k=9\times 10^9Nm^2/c^2 \\ \\ q_{1}=0.00150 C \ and \ q_{2}=0.00240 C \\ \\ r=0.900 m$

Then:

$F=9\times 10^9\frac{\left | 0.00150 \times 0.00240 \right |}{(0.900)^2} \\ \\ \therefore \boxed{F=40000N}$

This force is repulsive because the two charges are positive and recall that two positive charges or two negative charges repel each other while a positive charge  and a negative charge attract each other.

C. Answer:

From the statement, we have two charged objects. Let's say that this charges are:

$q_{1} \ and \ q_{2}$

If the amount of charge on one of the objects is tripled, let's say this is the charge $q_{2}$, then the new charge is:

$q_{N}=3q_{2}$

In the formula of Coulomb:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2} \\ \\ \therefore F=k\frac{\left | q_{1}(3q_{2}) \right |}{r^2} \\ \\ \therefore \boxed{F=3k\frac{\left | q_{1}q_{2} \right |}{r^2}}$

The conclusion is that if the amount of charge on one of the objects is tripled, the electric force between two charged objects is also tripled

d. Answer:

Let's use the Coulomb's Law again to solve this problem. We want to know how the electric force between two charged objects changes if the charges are moved closer together:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2}$

By saying that the charges are moved closer together, we want to express that r becomes smaller. Since r is in the denominator, this implies that the electric force between these two charged objects becomes greater.

e. Answer:

From the figure, we can see a metal sphere on a stand. There we have both positive and negative charges. We can say that the positive charge of this sphere is +10q and the negative and the negative charge is -10q. Since the electric charge is conserved, then the algebraic sum of all the electric charges in any closed system is constant. In conclusion, the sphere has no net charge.

f. Answer:

Here we want to know how the negative charges in the same sphere are redistributed when a positively charged rod is brought near it. Therefore, positive charge on rod  repels positive charges on the sphere, creating  zones of negative and  positive charge as indicated in the second Figure.

Answer 2

a) See the figure

b) According to Coulomb's law:

$F=\frac{1}{4\pi \epsilon_0}\cdot \frac{q_1q_2}{r^2}= 39944.67$ N

c) As above, the force will triple.

d) According to Coulomb's law the force will increase cause r will shorten.

e) Calculate charges of each sign: 10 positive and 10 negative. Total charge thus is 0.

f) See the figure. Positive repel from each other, positive and negative attract.