Question

Two-point charges q1 and q2 are held 4.00 cm apart vertically. An electron released at the middle point that is equidistant from both charges undergoes an initial acceleration of 8.95 X 10^18 m/s2 directly upward, parallel to the line connecting q1 and q2. Find the magnitude and direction of q1 and q2

Answer 1

Answer:

2.55*10^5 C

Explanation:

The forces due to q1 and q2 charges, accelerates the electron that is in the middle of the distance between the charges. That is

$F_{q1}+F_{q2}=m_ea_e$

If the electron is accelerated upward. Hence we have

$\frac{q_1e}{4\pi \epsilon_0 r^2}-\frac{q_2e}{4\pi \epsilon_0 r^2}=m_ea_e\\\\\frac{e}{4\pi \epsilon_0 r^2}(q_1-q_2)=m_ea_e\\\\\frac{1.6*10^{-19}C}{4\pi (0.02m)^2}(q_1+q_2)=(9.1*10^{-31}kg)(8.95*10^{18}\frac{m}{s^2})=8.14*10^{-12}N\\(q_1-q_2)=2.55*10^5C$

where we have taken r=2cm=0.02m; mass of electron=9.1*10^-31kg

charge of the electron=1.6*10^-19C

Hope this helps!!