Question

3 Which of the following is NOT correct?
Electric charge cannot be produced by frictional forces
Electric charge is conserved
Electric charges are integral multiple of the electronic charge
induced charges are the result of charge separation within a material

4 Which of the following is not true of an electrostatic force?
it acts along the line joining the point charges
it obeys an inverse square law
it is mutually attractive or repulsive
it is weaker than the gravitational attraction between the charged particles.

5 A tiny ball of mass 0.60 g is suspended from a rigid support with a piece of thread in a horizontal electric field of intensity 700 N/C. The ball is in equilibrium when the thread is inclined at an angle of
20o
to the vertical. What are the magnitude and sign of the charge on the ball? Take
g=9.8m/s2
−3.1×10−6C
3.2×10−6C
4.2×10−6C
−4.1×10−3

Answer 1

Answer on Question 62094, Physics, Electromagnetism

Question:

3) Which of the following is not correct?

a) electric charge cannot be produced by frictional forces

b) electric charge is conserved

c) electric charges are integral multiple of the electronic charge

d) induced charges are the result of charge separation within a material

Answer:

Electric charge can be produced by friction. For example, if we comb the hair, the electrons leave the hair and pass to the comb. Therefore, hair becomes positively charged and the comb – negatively charged. Thus, the incorrect answer is a) electric charge cannot be produced by frictional forces.

4) Which of the following is not true of an electrostatic force?

a) it acts along the line joining the point charges

b) it obeys an inverse square law

c) it is mutually attractive or repulsive

d) it is weaker than the gravitational attraction between the charged particles

Answer:

If we consider an atomic and subatomic levels (charged particles like electrons or protons, for instance) than the electrostatic force will be greater than the gravitational attraction between the charged particles. For example, for the electron and proton, the gravitational force is 39 orders of magnitude weaker than the electrostatic force. Therefore, the incorrect answer is d) it is weaker than the gravitational attraction between the charged particles.

5) A tiny ball of mass $0.60g$ is suspended from a rigid support with a piece of thread in a horizontal electric field of intensity $700N / C$ . The ball is in equilibrium when the thread is inclined at an angle of $20{}^{\circ}$ to the vertical. What are the magnitude and sign of the charge on the ball? Take $g = 9.8ms^{-2}$ .

a) $-3.1\cdot 10^{-6}C$

b) $3.2 \cdot 10^{-6} C$

c) $4.2 \cdot 10^{-6} C$

d) $-4.1 \cdot 10^{-3} C$

Solution:

Let's draw a free-body diagram:

Here, $W = mg$ is the force of gravity (or weight of the tiny ball), $F_{e} = qE$ is the electric force that acts on the tiny ball, $T$ is the force of tension in the thread, $T_{x}, T_{y}$ is the projections of the force of tension on axis $x$ and $y$ , respectively, $\theta$ is the angle of inclination of the thread to the vertical, $q$ is the charge of the tiny ball and $E$ is the intensity of the electric field.

Let's write the conditions of the equilibrium for the tiny ball:

$$\sum F _ {x} = 0, \quad \sum F _ {y} = 0.$$

Let's consider the forces that act on the tiny ball in the horizontal $x$- and vertical $y$-direction:

$$\sum F _ {x} = - T \sin \theta + F _ {e} = - T \sin \theta + q E = 0,$$ $$\sum F _ {y} = T \cos \theta - m g = 0.$$

We can express the force $T$ from the second equation:

$$T = \frac {m g}{\cos \theta}.$$

Then we can substitute it into the first equation and get:

$$- \frac {m g}{\cos \theta} \sin \theta + q E = 0,$$ $$- m g \tan \theta + q E = 0.$$

From the last equation we can find the magnitude of the charge on the ball:

$$q E = m g \tan \theta ,$$ $$q = \frac {m g \tan \theta}{E} = \frac {0 . 6 \cdot 1 0 ^ {- 3} k g \cdot 9 . 8 \frac {m}{s ^ {2}} \cdot \tan 2 0 {}^ {\circ}}{7 0 0 \frac {N}{C}} = 3. 0 6 \cdot 1 0 ^ {- 6} C \sim 3. 1 \cdot 1 0 ^ {- 6} C.$$

Let's discuss about the charge of the ball. Let's suppose that the electric field directed in the positive $x$-direction. Then, if we charged the ball with positive charge, the ball inclined at an angle $20{}^{\circ}$ to the vertical to the right. Now, if we charged the ball with negative charge, the ball inclined at an angle $20{}^{\circ}$ to the vertical in the opposite direction to the left. Now, let us imagine that we change the direction of the electric field. In this case, everything will be contrary. The positively charged ball will incline to the left while negatively charged ball will incline in opposite direction to the right. Therefore, both answers about the charge of the ball – either positive or negative are correct.

Answer:

a) $-3.1\cdot 10^{-6}C$

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