Question

charges q1=0.09 C, q2=0.01 C, are a distance /=1m apart. a charge Q is held fixed on the line between them, a distance x from q1. what value must Q,x have for q1,q2 to feel no net force?

Answer 1

Answer to Question #67118, Physics / Electromagnetism

Problem: Charges q1=0.09 C, q2=0.01 C, are a distance l = 1m apart. a charge Q is held fixed on the line between them, a distance x from q1. What value must Q,x have for q1,q2 to feel no net force?

Solution:

For the charges q1 and q2 to feel no net force, the value of electric field at the charges should be equal to 0.

Considering the picture, one can write

$$\left\{ \begin{array}{l} E _ {2} = E _ {Q 1} \\ E _ {1} = E _ {Q 2} \end{array} \right.$$

Where $E_2$ is the field of $q_2$ at the point of $q_1$, $E_{Q1}$ is the field of $Q$ at the point of $q_1$, $E_1$ is the field of $q_1$ at the point of $q_2$ and $Eq_2$ is the field of $Q$ at $q_2$.

Considering that in general $E = kq / d^2$ one can write the following:

$$\frac {k q _ {2}}{l ^ {2}} = \frac {k Q}{x ^ {2}}$$

And

$$\frac {k q _ {1}}{l ^ {2}} = \frac {k Q}{(l - x) ^ {2}}$$

From (1):

$$Q = q _ {2} * \frac {x ^ {2}}{l ^ {2}}$$

Then

$$\frac {k q _ {1}}{l ^ {2}} = \frac {k q _ {2}}{(l - x) ^ {2}} * \frac {x ^ {2}}{l ^ {2}}$$ $$q _ {1} (l ^ {2} - 2 x l + x ^ {2}) = q _ {2} x ^ {2}$$ $$(q _ {1} - q _ {2}) x ^ {2} - 2 l q _ {1} x + q _ {1} l ^ {2} = 0$$ $$x = \frac {2 q _ {1} l \pm \sqrt {(4 q _ {1} l ^ {2} - 4 * (q _ {1} - q _ {2}) q _ {1} l ^ {2})}}{2 (q _ {1} - q _ {2})}$$ $$x = \left\{ \begin{array}{l l} 1.5m & (1) \\ \frac {3}{4} m & (2) \end{array} \right.$$

The first solution is not between the charges, so we chose

$$x = \frac {3}{4} m$$

Then

$$Q = q _ {2} * \frac {x ^ {2}}{l ^ {2}} = 0.005625C$$

Answer:

$$x = \frac {3}{4} m, Q = q _ {2} * \frac {x ^ {2}}{l ^ {2}} = 0.005625C$$

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