Question 2 Consider two charges, where one (+3.80nC) is at the origin and the other (-14.4nC) is at the position x=2.86 86 cm. Find the xcoordinate where a proton would experience zero net force. Answer: square

Here's how to find the x-coordinate where a proton would experience zero net force:<br /><br />**1. Understand the Problem**<br /><br />A proton placed between two charges of opposite signs will experience two forces in opposite directions. We need to find the point where these forces balance each other out.<br /><br />**2. Coulomb's Law**<br /><br />The force between two charges is given by Coulomb's Law:<br /><br />F = k * |q1 * q2| / r^2<br /><br />Where:<br /><br />* F is the force<br />* k is Coulomb's constant (8.99 x 10^9 N m^2/C^2)<br />* q1 and q2 are the charges<br />* r is the distance between the charges<br /><br />**3. Set up the Equation**<br /><br />Let x be the distance from the origin where the proton experiences zero net force. The distance from the positive charge (+3.80 nC) to the proton is x. The distance from the negative charge (-14.4 nC) to the proton is (2.86 cm - x), or (0.0286 m - x).<br /><br />The force from the positive charge on the proton (q_p) is:<br /><br />F1 = k * |q1 * q_p| / x^2<br /><br />The force from the negative charge on the proton is:<br /><br />F2 = k * |q2 * q_p| / (0.0286 - x)^2<br /><br />Since the net force is zero, F1 = F2:<br /><br />k * |q1 * q_p| / x^2 = k * |q2 * q_p| / (0.0286 - x)^2<br /><br />**4. Simplify and Solve**<br /><br />Notice that k and q_p cancel out:<br /><br />|q1| / x^2 = |q2| / (0.0286 - x)^2<br /><br />Substitute the values of q1 and q2:<br /><br />3.80 x 10^-9 / x^2 = 14.4 x 10^-9 / (0.0286 - x)^2<br /><br />Simplify further:<br /><br />3.80 / x^2 = 14.4 / (0.0286 - x)^2<br /><br />Cross-multiply:<br /><br />3.80 * (0.0286 - x)^2 = 14.4 * x^2<br /><br />Take the square root of both sides:<br /><br />√(3.80/14.4) * (0.0286-x) = x<br />0.515 * (0.0286 - x) = x<br />0.0147 - 0.515x = x<br />0.0147 = 1.515x<br />x ≈ 0.0097 m<br /><br />**5. Convert to cm**<br /><br />x ≈ 0.0097 m * 100 cm/m = 0.97 cm<br /><br />**Answer:** 0.97 cm (approximately)<br />