Question

HELP ME PLEASE!!!!!!!!!

A 2 µC charge q1 and a 2 µC charge q2 are 0.3 m from the x-axis. A 4 µC charge q3 is 0.4 m from the y-axis. The distances d13 and d23 are 0.5 m. Find the magnitude and direction of the resulting vector R. Round your answer to the nearest tenth.

Answer 1

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

$F = \frac{kq_1q_3}{d_{13}^2}$

now from the above equation

$F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}$

$F = 0.288 N$

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

$F_{net} = 2Fcos\theta$

here we know that angle is

$\theta = 37 degree$

now we have

$F_{net} = 2\times 0.288 cos37$

$F_{net} = 0.46 N$

so net force on q3 is 0.46 N vertically upwards along +Y axis

Answer 2

Answer:

Explanation:

E2020