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 cm. Find the xcoordinate where a proton would experience zero net force. Answer: square disappointed

Here's how to find the x-coordinate where a proton would experience zero net force:<br /><br />**1. Understand the Problem**<br /><br />We have two charges: a positive charge (q1) at the origin and a negative charge (q2) at x = 2.86 cm. A proton (positive charge) will be repelled by the positive charge and attracted by the negative charge. We need to find the point on the x-axis where these two forces balance each other out.<br /><br />**2. Set up the Equation**<br /><br />Let the position of the proton be x. The force exerted on the proton by q1 is:<br /><br />F1 = k * q1 * qp / x^2<br /><br />where:<br />* k is Coulomb's constant (8.99 x 10^9 N m^2/C^2)<br />* q1 is the first charge (+3.80 nC = 3.80 x 10^-9 C)<br />* qp is the charge of the proton (1.60 x 10^-19 C)<br />* x is the distance between the proton and q1<br /><br />The force exerted on the proton by q2 is:<br /><br />F2 = k * q2 * qp / (2.86 cm - x)^2 = k * q2 * qp / (0.0286 m - x)^2<br /><br />where:<br />* q2 is the second charge (-14.4 nC = -14.4 x 10^-9 C)<br />* 0.0286 m - x is the distance between the proton and q2 (converted cm to meters)<br /><br />For the net force to be zero, F1 and F2 must be equal in magnitude:<br /><br />F1 = F2<br /><br />k * q1 * qp / x^2 = k * |q2| * qp / (0.0286 - x)^2<br /><br />**3. Solve for x**<br /><br />Notice that k and qp 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 and 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 />sqrt(3.80) * (0.0286 - x) = sqrt(14.4) * x<br /><br />1.949 * (0.0286 - x) = 3.795 * x<br /><br />0.0558 - 1.949x = 3.795x<br /><br />0.0558 = 5.744x<br /><br />x = 0.0558 / 5.744<br /><br />x ≈ 0.00971 m = 0.971 cm<br /><br />Since the negative charge is larger in magnitude, the point of zero force must be to the left of the positive charge (closer to the origin). Therefore, the x-coordinate is approximately -0.971 cm.<br /><br />**Answer:** -0.971 cm<br />