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Electric fields describe the interaction of stationary charged matter. They underlie the working of diverse technology from atom smashers to the poor cell reception you're getting right now.

As shown in the above figure, the point charges \( q_A = 4 \text{ C} \) and \( q_B \) produce potential \( V \) at the point \( X, \) which is away from \( q_A \) by a distance of \( d \) and away from \( q_B\) by a distance of \( 2d.\) If the point charge \( q_C \) is away from \( X\) by a distance of \( d \) and the total potential at \(X \) is \( -3V, \) what is the sum of the point charges \( q_A, \) \( q_B \) and \( q_C?\)

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There are two charged metal spheres, as shown in the above figure. The sphere \( A \) with radius \( r_1 =16 \text{ cm} \) is charged with charge \( Q_1 =32 \text{ C} ,\) and the sphere \( B \) with radius \( r_2 =8 \text{ cm} \) is charged with charge \( Q_2 = 4 \text{ C}. \) If the two spheres touch each other, what is the final charge of the sphere \( A? \)

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Three positively charged particles lie on a straight line. Their charges are \(q_1= 4 \text{ C}, q_2 =5 \text{ C} \) and \(q_3=16 \text{ C},\) with positions \(x_1 = 0, \) \(x_2\) and \(x_3 = r = 24 \text{ m}\). The equilibrium point is placed between \( q_1\) and \( q_3. \)

When \(q_1\) and \(q_3 \) are fixed, and \(q_2 \) is movable, find the equilibrium point for \( q_2. \)

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As shown in the above diagram, there are three electrically neutral metal spheres: metal sphere \(A \) with radius \( r_A=4 \text{ cm}, \) metal sphere \(B \) with radius \( r_B = 2 \text{ cm} \) which is positively charged with charge \( q_B = 6.0 \text{ C} ,\) and metal sphere \( C \) with radius \( r_C = 6 \text{ cm} \) which is negatively charged with charge \( q_C = -10.0 \text{ C}. \) If \( B \) touches \( A \) and then touches \( C, \) what is the final charge of sphere \( B?\)

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