5. A charge of -5.00 nC and a charge of -2.00 nC are separated by a distance of 40.0 cm. Find the equilibrium position (where the net force equals zero) for a third charge of 15.0 nC.
Sheeeeeeeesh
To find the equilibrium position for the third charge, we need to consider the forces exerted by the other two charges on it.
The force between two charges is given by Coulomb's law:
\[ F = \frac{{k \cdot |q_1 \cdot q_2|}}{{r^2}} \]
where F is the force, k is the electrostatic constant (9.0 × 10^9 N m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between them.
Let's assume that the distance between the third charge and the charge of -5.00 nC is x, and the distance between the third charge and the charge of -2.00 nC is 40 cm - x.
Now let's calculate the forces exerted by the two charges on the third charge:
The force by the charge of -5.00 nC on the third charge is given by:
\[ F_1 = \frac{{k \cdot |q_1 \cdot q_3|}}{{x^2}} \]
The force by the charge of -2.00 nC on the third charge is given by:
\[ F_2 = \frac{{k \cdot |q_2 \cdot q_3|}}{{(40 \, \text{cm} - x)^2}} \]
At equilibrium, the net force on the third charge is zero. Therefore, \( F_1 + F_2 = 0 \).
Substituting the values of q1, q2, q3, k, and r into the above equation, we get:
\[ \frac{{k \cdot |q_1 \cdot q_3|}}{{x^2}} + \frac{{k \cdot |q_2 \cdot q_3|}}{{(40 \, \text{cm} - x)^2}} = 0 \]
Simplifying the equation, we find:
\[ |q_1 \cdot q_3| + |q_2 \cdot q_3| = 0 \]
Substituting the values of q1, q2, and q3 into the above equation, we get:
\[ |-5.00 \, \text{nC} \cdot 15.0 \, \text{nC}| + |-2.00 \, \text{nC} \cdot 15.0 \, \text{nC}| = 0 \]
Simplifying further, we find:
\[ |-75.0 \, \text{nC}^2| + |-30.0 \, \text{nC}^2| = 0 \]
Since the absolute values of both terms are positive, the equation cannot be satisfied. Hence, there is no equilibrium position for the third charge with these initial conditions.