Question

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

a) −3.1×10−6C

b)3.2×10−6C

c)4.2×10−6C

d)−4.1×10−3

6) The following are true about electric field lines except that they
a) are drawn such that the magnitude of the field is proportional to the number of lines crossing a unit area perpendicular to the lines
b) do not intersect one another
c) are discontinuous and may terminate in a vacuum
d) give the direction of motion of a unit positive test-charge under the action of the electrostatic force

Answer 1

Answer on Question #61327, Physics / Electromagnetism

5) A tiny ball of mass $0.60\mathrm{g}$ is suspended from a rigid support with a piece of thread in a horizontal electric field of intensity $700\mathrm{N/C}$. 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 $\mathrm{g} = 9.8\mathrm{m/s}^2$

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

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

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

d) $-4.1 \times 10^{-3}$

Find: $q - ?$

Given:

$$m = 0.6 \times 10^{-3} \mathrm{kg}$$ $$E = 700 \mathrm{N/C}$$ $$\alpha = 20{}^{\circ}$$ $$g = 9.8 \mathrm{m/s}^2$$

Solution:

Consider the forces which acting on the tiny ball $q$.

Newton's Second Law:

$$\vec{F} = m \vec{a} \quad (1)$$ $$\text{Of} \quad (1) \Rightarrow \vec{T} + m \vec{g} + \vec{f} = m \vec{a} \quad (2),$$

where $\vec{T}$ is tension force,

$m\vec{g}$ is gravity,

$\vec{f}$ is force of electric field

Projections of the vectors:

OX: $-T \sin \alpha + f = 0$ (3)

OY: $T \cos \alpha - mg = 0$ (4)

Force of electric field:

$f = E|q|$ (5)

(5) in (3): $T \sin \alpha = E|q|$ (6)

Of (4) $\Rightarrow T \cos \alpha = mg$ (7)

We divide (6) on (7) term by term:

$$\tan \alpha = \frac{E|q|}{mg} \quad (8)$$

Of (8) $\Rightarrow |q| = \frac{mg \tan \alpha}{E}$ (9)

Of (9) $\Rightarrow |q| = 3.1 \times 10^{-6} C$

From Figure $\Rightarrow$ sign of the charge: $q = -3.1 \times 10^{-6} \, \text{C}$

**Answer:**

a) $-3.1 \times 10^{-6} \, \text{C}$

6) The following are true about electric field lines except that they

a) are drawn such that the magnitude of the field is proportional to the number of lines crossing a unit area perpendicular to the lines

b) do not intersect one another

c) are discontinuous and may terminate in a vacuum

d) give the direction of motion of a unit positive test-charge under the action of the electrostatic force

**Answer:**

c) are discontinuous and may terminate in a vacuum

http://www.AssignmentExpert.com