3. [-/2 Points] DETAILS SERPSE10 22.3.OP.003. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Two charged particles, q, and q2, are located on the x-axis, with q, at the origin and q2 initially at x] = 13.2 mm. In this configuration, q, exerts a repulsive force of 2.62 UN on 92. Particle q2 is then moved to X2 = 17.0 mm. What is the force (magnitude and direction) that q, exerts on q, at this new location? (Give the magnitude in UN.) magnitude direction --Select-- v Need Help? Read It Submit Answer 4. [-/4 Points] DETAILS SERPSE10 22.3.OP.004. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER (a) Two protons in a molecule are 4.20 x 10-10 m apart. Find the electric force exerted by one proton on the other. magnitude N direction ---Select--- V (b) State how the magnitude of this force compares with the magnitude of the gravitational force exerted by one proton on the other. (c) What must be a particle's charge-to-mass ratio if the magnitude of the gravitational force between two of these particles is equal to the magnitude of electric force between them? C/kg Need Help? Read It Watch It Fwer Screenshot 5. [-/2 Points] DETAILS SERPSE10 22.3.OP.007.MI. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Three point charges are arranged as shown in the figure below. Find the magnitude and direction of the electric force on the particle q = 5.28 nC at the origin. (Let 12 = 0.245 m.) magnitude IN direction counterclockwise from the +x axis 6.00 nC 0.100 m -3.00 nC Need Help? Read It Master It6. [-/10 Points] DETAILS SERPSE10 22.AE.003. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Where Is the Net Force Zero? Three point charges lie along the x-axis as shown in the figure. The positive charge q, = 19.0 JC is at x = 2.00 m, the positive charge q, = 9.00 UC is at the origin, and the net force acting on 43 is zero. What is the x-coordinate of 93? Three point charges are placed along the x-axis. If the net force on q; is zero, the force F 13 exerted by q, on 4; must be equal in magnitude and opposite in direction to the force F 23 exerted by 42 on 43- - 2.00 m 2.00 - x- + + 92 43 41 SOLUTION Conceptualize Because q; is near two other charges, it experiences two electric forces. The forces lie along the same line in this problem as indicated in the figure. Because q, is negative and 41 and 42 are positive, the forces Fig and F23 are both attractive. Because q2 is the smaller charge, the position of q; at which the force is zero should be |-Select-- v q2 than q,. Categorize Because the net force on q; is zero, we model the point charge as a particle -Select-- Screenshot Use the following as necessary: x.) Write an expression for the net force on charge q; when it is in equilibrium: 19, 11931 F3 = F23 + F 13 = -Ke-2 * 1921931 + ke x 2 i = 0 Move the second term to the right side of the equation and set the coefficients of the unit vector i equal: 19, 11931 19211931 x2 Eliminate k, and |q; | and rearrange the equation: ) 1921 = * 21911Move the second term to the right side of the equation and set the coefficients of the unit vector i equal: 19, 11931 19211931 Eliminate k, and |q; | and rearrange the equation: ) 1921 = *21911 Take the square root of both sides of the equation (enter the positive root): V1921 = 1 XV 1911 Solve for x: V1921 X V 1921 = V 19,1 Substitute numerical values, choosing the plus sign. (Calculate x in m.) X = m Finalize Notice that the movable charge is indeed closer to q, as we predicted in the Conceptualize step. The second root to the quadratic equation (if we choose the negative sign) is x = -4.42 m. That is another location where the magnitudes of the forces on q, are equal, but both forces are in | ---Select--- v , so they do not cancel. EXERCISE Two charges, q, = -10.0 UC and q2 = 50.0 UC are located on the x-axis. Charge q, is at the origin, and q2 is at x = 10.0 cm. Where must a positive charge q be placed (on the x-axis, in cm) en that the net force on q is zero? Screenshot em Need Help? Read It