Answer to Question #160199 in Physics for Smayya Ruganyira

Question

TRUE or FALSE

a) An object which is positively charged contains all protons and no electrons.

b) An object which is negatively charged could contain only electrons with no accompanying protons.

c) An object which is electrically neutral contains only neutrons.

Draw the electric field pattern lines between:

A. Two equal positive point charges.

B . two equal negative point charges.

Two insulated metal spheres carrying charges of +6nC and −10nC are separated by a distance of 20 mm.

A. What is the electrostatic force between the spheres?

B. The two spheres are touched and then separated by a distance of 60mm. What are the new charges on the spheres?

C. What is new electrostatic force between the spheres at this distance?

The electrostatic force between two charged spheres of +3nC and +4nC respectively is 0.04N. What is the distance between the spheres?

Two particles having charges of 0.70nC and 12nC are separated by a distance of 2m. At what point along the line connecting the two charges is the net electric field due to the two charges equal to zero? 

Calculate the potential difference between two parallel plates if it takes 5000J of energy to move 25C of charge between the plates?

Calculate the electric field between the plates of a capacitor if the plates are 20mm apart and the potential difference between the plates is 300V.

Calculate the electrical potential energy of a 6nC charge that is 20cm from a 10nC charge.

Answer 1

a) False

b) False

c) False

A)

B)

a)

$$F_e = \frac{k Q_1 Q_2}{r^2}= \frac{(9\cdot10^9)(-10\cdot10^{-9})(6\cdot10^{-9})}{0.02^2}=-0.00135\ N$$

b)

$$Q=(6-10)=-4\ nC$$

c)

$$F_e = \frac{k Q_1 Q_2}{r^2}= \frac{(9\cdot10^9)(-4\cdot10^{-9})(-4\cdot10^{-9})}{0.02^2}=0.00036\ N$$

1)

$$F_e = \frac{k Q_1 Q_2}{r^2}\\ 0.04= \frac{(9\cdot10^9)(3\cdot10^{-9})(4\cdot10^{-9})}{r^2}\\r=0.0016\ m$$

2)

$$\frac{(9\cdot10^9)(0.7\cdot10^{-9})}{x^2}=\frac{(9\cdot10^9)(12\cdot10^{-9})}{(2-x)^2}\\x=0.389\ m$$

3)

$$U=\frac{5000}{25}=200\ V$$

4)

$$E=\frac{300}{0.02}=15000\frac{V}{m}$$

5)

$$E_p= \frac{(9\cdot10^9)(6\cdot10^{-9})(10\cdot10^{-9})}{0.2}=2.7\cdot10^{-6}\ J$$