Question

A charge of 0.8 coulomb is divided into two charges $Q_1$ and $Q_2$. These are kept at a separation of 30cm. The force on $Q_1$ is maximum when

• A. $Q_1 = Q_2 = 0.4C$
• B. $Q_1 = 0.8C$, $Q_2$ negligible
• C. $Q_1$ negligible, $Q_2 = 0.8C$
• D. $Q_1 = 0.2C$, $Q_2 = 0.6C$

Answer 1

Answer: (A)

$Q_1 = Q_2 = 0.4C$

Answer 2

Let the total charge be $Q = 0.8$ C. This charge is divided into two parts, $Q_1$ and $Q_2$.
So, $Q_1 + Q_2 = Q = 0.8$ C.
These charges are kept at a separation of $r = 30$ cm.
The magnitude of the force between the two charges is given by Coulomb's Law:
$F = k \frac{|Q_1 Q_2|}{r^2}$, where $k$ is Coulomb's constant.
The separation $r$ is fixed, and $k$ is a constant. To maximize the force $F$, we need to maximize the product $|Q_1 Q_2|$.

Since the total charge is positive (0.8 C) and the options suggest that the two parts are positive, we assume $Q_1 \ge 0$ and $Q_2 \ge 0$. In this case, we need to maximize the product $Q_1 Q_2$.

We have the constraint $Q_1 + Q_2 = 0.8$.
Let $Q_1 = x$. Then $Q_2 = 0.8 - x$.
We want to maximize the product $P(x) = Q_1 Q_2 = x(0.8 - x)$.
$P(x) = 0.8x - x^2$.

To find the maximum value of $P(x)$, we can use calculus. Differentiate $P(x)$ with respect to $x$ and set the derivative to zero:
$\frac{dP}{dx} = \frac{d}{dx}(0.8x - x^2) = 0.8 - 2x$.
Set $\frac{dP}{dx} = 0$:
$0.8 - 2x = 0$
$2x = 0.8$
$x = 0.4$.

To confirm that this is a maximum, we can check the second derivative:
$\frac{d^2P}{dx^2} = \frac{d}{dx}(0.8 - 2x) = -2$.
Since the second derivative is negative, the value $x = 0.4$ corresponds to a maximum.

So, the product $Q_1 Q_2$ is maximum when $Q_1 = 0.4$ C.
If $Q_1 = 0.4$ C, then $Q_2 = 0.8 - Q_1 = 0.8 - 0.4 = 0.4$ C.

Thus, the force on $Q_1$ (and also on $Q_2$) is maximum when the total charge is divided equally into two parts: $Q_1 = 0.4$ C and $Q_2 = 0.4$ C.