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

2.55*10^5 C

Step-by-step 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

`(q_1e)/(4\pi \epsilon_0 r^2)-(q_2e)/(4\pi \epsilon_0 r^2)=m_ea_e\\\\(e)/(4\pi \epsilon_0 r^2)(q_1-q_2)=m_ea_e\\\\(1.6*10^(-19)C)/(4\pi (0.02m)^2)(q_1+q_2)=(9.1*10^(-31)kg)(8.95*10^(18)(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!!