The force between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by Coulomb's law: \[ F = k \frac{|q_1 q_2|}{r^2} \] where: - \( k \) is Coulomb's constant,
- \( r \) is the distance between the charges.
For the charges on spheres A and B, we are given:
- \( q_1 = 6 \, {C} \),
- \( q_2 = 12 \, {C} \),
- The force between them is \( F \).
Now, we are asked to find the charge \( q_1' \) that must be given to sphere A in order to reverse the direction of the force to \( -F \).
The charges should attract each other instead of repelling, meaning the product of the charges should become negative.
To reverse the force direction, the charge \( q_1' \) on sphere A must be: \[ q_1' = -12 \, {C} \] Thus, the charge to be given to sphere A to reverse the force is \( -12 \, {C} \). Therefore, the correct answer is: \[ \boxed{-12 \, {C}} \]
\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]
Then \( f'(-4) \) is equal to:If \( f'(x) = 4x\cos^2(x) \sin\left(\frac{x}{4}\right) \), then \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is equal to:
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
\[ \int \frac{4x \cos \left( \sqrt{4x^2 + 7} \right)}{\sqrt{4x^2 + 7}} \, dx \]