Question

QUESTION 4 In the diagram below, if q 2 is positive then what sign is the charge on q 1 in order to produce the force required to cause the deflection in the string that supports q 1? 91 F21 - O Not enough information to decide. Positive Neutral O Negative QUESTION 5 in the question abo charge q 2 s 1.7x10-9C and charge q 1 has mass 024g The separation r is 3 3cm and the angle 0 is 16.7 degrees. Find q 1 magnitude and s this question is very small and using scientific notation will help. This system uses E in place of x10 so for example to enter 3x10-24 type 3E 24 in the answer box you don't need to enter a . for pesi eansvers but you wil need to enter a t

question 5. referencing question 4.

Answer 1

5)

Forces acting on the charge $q_1$ are

i) Weight $F_g=m_1g$ ,vertically downwards

ii) Force of repulsion between the two charges is $F_{21}=\frac{q_1q_2}{4\pi\epsilon_0 r^2}$ and is directed horizontally to  left.

iii)Components of tension in the string $T\cos\theta$ vertically up, $T\sin\theta$ horizontally to right.

Net vertical force acting on the charge $q_1$ is zero, $T\cos\theta-F_g=0$$\Rightarrow T\cos\theta=m_1g$---(1)

Net horizontal force acting on the charge $q_1$ is zero, $F_{21}-T\sin\theta=0\Rightarrow T\sin\theta=\frac{q_1q_2}{4\pi\epsilon_0 r^2}$ ----(2)

Dividing equations (2) by (1) $\tan\theta=\frac{q_1q_2}{4\pi\epsilon_0 r^2m_1g}$

$q_1=\frac{4\pi\epsilon_0\,\tan\theta r^2m_1g}{q_2}=\frac{4\pi*8.85*10^{-12}\tan16.7\degree*(3.3*10^{-2})^2*(0.24*10^{-3})*9.81}{1.7*10^{-9}}$

$q_1=50.3*10^{-9}\,C$

q₁ = 50.3 E-9