Question

The drawing shows an equilateral triangle, each side of which has a length of 2.00 cm. Point charges are fixed to each corner, as shown. The 4.00 microC charge experiences a net force due to the charges qA and q B. This net force points vertically downward and has a magnitude of 405 N. Determine the magnitudes and algebraic signs of the charges qA and q B.

Answer 1

Answer:

$-2.60097\times 10^{-6}\ C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

r = Distance between charges = 2 cm

The electric force is given by

$F=\dfrac{k4\times 10^{-6}q_A}{r^2}$

Also

$F=\dfrac{k4\times 10^{-6}q_B}{r^2}$

The sum of the force is given by

$\sum F=\dfrac{k4\times 10^{-6}q_A}{r^2}cos 30+\dfrac{k4\times 10^{-6}q_B}{r^2}cos 30\\\Rightarrow \sum F=2\dfrac{k4\times 10^{-6}q_A}{r^2}cos 30\\\Rightarrow q_a=\dfrac{\sum Fr^2}{2\times k4\times 10^{-6}cos30}\\\Rightarrow q_a=\dfrac{405\times 0.02^2}{2\times 8.99\times 10^9\times 4\times 10^{-6}cos30}\\\Rightarrow q_a=2.60097\times 10^{-6}\ C$

Thus $q_a=q_b=-2.60097\times 10^{-6}\ C$.

The negative sign is because the force points downward which is taken as negative

Answer 2

The magnitude of each reference charge  $q_{A}$ and $q_{B}$ is $+4.5\;\rm \mu C$. And their algebraic signs are positive.

Given data:

The length of side of equilateral triangle is, L = 2.00 cm.

The magnitude of test charge is, $q= 4.00\;\rm \mu C = 4.00 \times 10^{-6} \;\rm C$.

The magnitude of net force is, $F = 405 \;\rm N$.

Let us consider two reference charges, say $q_{A}$ and $q_{B}$. Then the force between the each of reference charge and test charge is,

$F = \dfrac{k \times q \times q_{A}}{L^{2}}$

here, k is the Coulomb's constant.

Solving as,

$405 = \dfrac{9 \times 10^{9} \times 4.00 \times 10^{-6} \times q_{A}}{0.02^{2}}\\\\q_{A}=4.5 \times 10^{-6} \;\rm C = 4.5\;\rm \mu C$

Similarly, solve for the value other reference charge as,

$F = \dfrac{k \times q \times q_{B}}{L^{2}}\\405 = \dfrac{9 \times 10^{9} \times 4.00 \times 10^{-6} \times q_{B}}{0.02^{2}}\\\\q_{B}=4.5 \times 10^{-6} \;\rm C = 4.5\;\rm \mu C$

Thus, we can conclude that the magnitude of each reference charge  $q_{A}$ and $q_{B}$ is $+4.5\;\rm \mu C$.

Learn more about the coulomb's force here:

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