✨ VIPSolutions

Let's solve the problem step by step. ## **Let us define:** - Two point charges \(q_1\) and \(q_2\) are separated by a distance \(d\). - We are looking for the place where the electric field is zero, specifically **one-third** of the way from \(q_1\) to \(q_2\). Let: - The position of \(q_1\) be at \(x=0\). - The position of \(q_2\) be at \(x=d\). - The point of interest is at \(x = \frac{d}{3}\) (one-third from \(q_1\) to \(q_2\)). ## **Electric field at \(x = \frac{d}{3}\):** - **Distance from \(q_1\):** \(r_1 = \frac{d}{3}\) - **Distance from \(q_2\):** \(r_2 = d - \frac{d}{3} = \frac{2d}{3}\) ### **Electric fields (magnitude):** - Field due to \(q_1\) at this point: \[ E_1 = k \frac{|q_1|}{\left(\frac{d}{3}\right)^2} = k \frac{9|q_1|}{d^2} \] - Field due to \(q_2\) at this point: \[ E_2 = k \frac{|q_2|}{\left(\frac{2d}{3}\right)^2} = k \frac{9|q_2|}{4d^2} \] ### **Field Zero Condition:** The fields must be **equal and opposite** at this point for net field to be zero. Assume both \(q_1\) and \(q_2\) are positive (charges have the same sign): - Both fields point away from their respective charges, so at this point, the fields are in **opposite directions** only if the charges have **opposite signs**. So, let's say \(q_1 > 0\) and \(q_2 < 0\). - The field due to \(q_1\) at \(\frac{d}{3}\) points **away from \(q_1\)** (to the right). - The field due to \(q_2\) at \(\frac{d}{3}\) (since \(q_2 < 0\)), the field points **toward \(q_2\)**, i.e., also to the right. But for the fields to **cancel**, one must point right and the other left. Let's clarify: - For an external point between two charges, the field will be zero only if the charges have **opposite signs**. Let's write the condition for zero net field at \(x = \frac{d}{3}\): \[ E_1 + E_2 = 0 \] But considering directions (let's assume right is positive): - \(E_1\) at \(x = \frac{d}{3}\) due to \(q_1\): rightward if \(q_1>0\). - \(E_2\) at \(x = \frac{d}{3}\) due to \(q_2\): toward \(q_2\). Since \(q_2\) is at \(x=d > \frac{d}{3}\), the field due to \(q_2\) at \(x=\frac{d}{3}\) is **leftward** if \(q_2 > 0\), rightward if \(q_2 < 0\). So, for the fields to cancel, we must have **charges of opposite sign**. Suppose \(q_1 > 0\), \(q_2 < 0\): - \(E_1\): rightward. - \(E_2\): rightward (since negative charge, field points toward \(q_2\)). They **add**, not cancel. So, that's not possible. Suppose \(q_1 > 0\), \(q_2 > 0\): - \(E_1\): rightward. - \(E_2\): leftward (since positive charge, field points away from \(q_2\)). So, they **oppose**. Now we can set \(E_1 = E_2\). ## **Calculate Ratio:** Set magnitudes equal for cancellation: \[ E_1 = E_2 \] \[ k \frac{q_1}{\left(\frac{d}{3}\right)^2} = k \frac{q_2}{\left(\frac{2d}{3}\right)^2} \] \[ \frac{q_1}{\frac{d^2}{9}} = \frac{q_2}{\frac{4d^2}{9}} \] \[ \frac{q_1}{1/9} = \frac{q_2}{4/9} \] \[ 9q_1 = \frac{9}{4}q_2 \] \[ 9q_1 = \frac{9}{4}q_2 \] \[ q_1 = \frac{1}{4}q_2 \] \[ \boxed{\frac{q_1}{q_2} = \frac{1}{4}} \] --- ## **Final Answer:** \[ \boxed{\frac{q_1}{q_2} = \frac{1}{4}} \] The ratio of \(q_1\) to \(q_2\) is **1:4**. (Both charges are positive.)

The electric field at a point between two charges is zero when the magnitudes of their fields are equal and opposite at that point. When the point is located a specific fraction along the line joining the charges, the ratio of their magnitudes depends on the inverse square law of electrostatics. For a point one-third of the way from \(q_1\) to \(q_2\), the ratio of the charges \(q_1:q_2\) is 1:4.