1. You have two isolated, equal and opposite point charges, -Q 20 apart, on a mo
ID: 1568481 • Letter: 1
Question
1. You have two isolated, equal and opposite point charges, -Q 20 apart, on a mounting board. On a straight line connecting the charges, at a point 15 cm away from the negative charge (Point A), the electric potential due to both charges is 4 1) Write the equation for the net potential of the two charges, +Q and -Q, at the specified point. 2) Find the magnitude Q of each charge. 3) Point B is also located on the line connecting the point charges, 5 cm away from the negative charge. Calculate the potential difference V 2. In the circuit shown below, R2 R 20 Ohms, the DC voltage source provides V 12.5 Volts, and the ammeter reads 0.50 A. 1) What is the value of RI? 2) What is the voltage drop across R2? 3) What is the current through R3? 3. You drop a magnet, North end down, through a horizontal conducting coil, and record the following EMF as a function of time: E emtv Image from http://www.a-levelphysicstutor.com/field-electro- mag-ind.php. l) Determine the direction of the induced magnetic field at theExplanation / Answer
1.1)
At the point A, potenital is positive, so A must be between the Q and -Q and, so the distance of A from Q is 5cm and from -Q is 15cm.
So the equation for potential at A is given by;
V = (1/40)[Q/(5cm) - Q/(15cm)] ---------(1)
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1.2)
In the equation (1) above put V = 4V;
4V = (1/40)[Q/(5cm) - Q/(15cm)] ---------- (2)
or, 4V = (9X109N-m2/C2)Q[1/0.05m - 1/0.15m]
or, Q = 3.33X10-11C
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1.3)
Point B is 5cm from -Q and 15cm from Q. So potential at B is,
VB = (1/40)[-Q/(5cm) + Q/(15cm)] ---- (here RHS is just the negative of the RHS of equation (2))
or, VB = -4V
So VA - VB = 4V - (-4V)
or, VA - VB = 8V
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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....
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