Question

1. a) The electric field at a certain point around a + 5.84C charge has a magnitude of 3.68 x 105N/C. What is the magnitude of the electrostatic force that would beexerted on a -3.72 C charge located at this point?

b) What is the direction of the electrostatic force referred to in (a)?

c) What is the direction of the electric field at the point referred to in (a)?

d) How far away from the + 5.84C charge is the -3.72C charge referred to in (a)?

2. Consider the electric field around a -4.88C charge. What would be the magnitude and direction of this field at a distance of 0.577 m from the -4.88 C charge?

Answer 1

First we will discuss what Electric field exactly is;

Electric field due to source charge at a point is defined as the force experienced per unit positive test charge placed at point without disturbing the source charge.

i.e Electric field, E is given by;

E = kq/r2 ...........eqn 1

where k = 1/4(pie)eo and has a value = 9*109 Nm2C-2  and r is the distance between the test charge and the source charge.

Q no.1 Part a Now, magnitude of source charge is given by, q = + 5.84 C

Positive sign indicates that it is positively charged and hence the electric field lines would be radially outwards.

Also, magnitude of Electric field due to this Charge at a certain point, E = 3.68*105 N/C

Using values of k, q and E in eqn 1, we can calculate the distance at which the the field value is 3.68*105 N/C

i.e 3.68*105 = 9*109 * 5.84/r2

or r2 = 9*109*5.84/3.68*105 = 52.56*109/3.68*105 = 14.282*104

or r = (14.282*104)1/2

r = 3.779 *102 m

Now, electrostatic force or Coulomb's force (F) between two charged particles q and q' is equal to product of magnitude of the two charges and inversely proportional to square of distance(r) between them.

i.e F = kqq'/r2 ..........eqn 2

where k is the constant of proportionality and has value = 9*109 Nm2C-2

According to question, we have placed a -ve charge, q' = -3.72C at distance r = 3.779*102 m away from the source charge q= 5.84 C, hence electrostatic force between them would be given by using eqn 2

F = 9*109 *(+5.84)*(-3.72)/ 3.779*102 = -195.5232*109/3.779*102 = -51.739*107 N

i.e F = -51.739*107 N ..........eqn 3

Eqn 3 gives the electrostatic force acting among two opposite charges placed at a distance of 3.779*102 m apart.

It must be noted that negative sign in eqn 3 indicates that the force is attractive in nature and that is what we expect exactly from two opposite charges i.e to attract each other.

Q no.1 Part b.

According to mentioned discussion above; The force is attractive in nature, i.e the source charge q would attract the charge q'= -3.72C towards itself; hence the direction of force will be inwards i.e towards the charge q.

Q no.1 Part c Electric field as mentioned by question is, E = 3.68*10⁵ N/C

the source charge is, q = +5.84C

For a positive charge, electric field lines are always pointed in radially outward direction. Hence, Electric field lines would be away from the charge q = +5.84 .One must not confuse the direction of Electric field lines with the direction of electrostatic force, because, Electric field lines are for a certain "POINT" that does not have any charge . However, electrostatic Force acts among two charges,hence both of them would either attract(for opposite charges) or repel (for similar charges)each other and direction governed accordingly.

Q no.1 Part d. From the diagram in part b of question 1,it is quite clear that distance between charge q=+5.84 C and q' = -3.72 C is r, which is equal to = 3.779*10²m

Q no.2 Charge q = -4.88C ..........eqn 4

The distance r where we want to calculate magnitude of Electric field , r = 0.577 m ......eqn 5

According to eqn 1 , E = kq/r² ,

Hence using values of k = 9*10⁹ Nm²C^-2, q and r from eqn 4 and 5 respectively in eqn1, we get

E = 9*109*(-4.88)/(0.577)2 = 9*10⁹(-4.88)/0.332929 = -43.92*109/ 0.332929 = - 131.92*109 NC-1

i.e E = -131.92*109 N/C

Above mentioned expression is that of  electric field due to a negative charge q= -4.88C at a distance of r =0.577 m  away from the charge. Since the magnitude of electric field is negative; i.e the source charge is negative in nature, this implies that electric field lines will be radially inwards i.e towards the charge q.