Two charges, +4µC on the left and -9µC on the right, are separated by 30cm. 1. W

Question

Two charges, +4µC on the left and -9µC on the right, are separated by 30cm.

1. What is the magnitude and direction of the electric force acting on the -9µC charge?





2. What is the magnitude and direction of the electric field measured at a point 10cm to the right of the +4µC charge?






3. At what point in space will the electric field due to these charges be zero?







4. What is the sign and magnitude of the electric potential at a point 10cm to the right of the +4µC charge?






5. How much work would it require to separate these two charges to an infinite distance?


6. A disposable camera has a 1.5-V battery with an internal resistance of 5000?. To prepare for usage of the flash, the battery charges a capacitor of 30mF.

a.) How long will it take the capacitor to charge up to 1.35V, 90% of the battery’s EMF?







b.) How much energy is stored in the capacitor when it holds the full 1.5V of the battery?







7. The shutter switch disconnects the capacitor from the battery, and places it across a flashbulb with resistance of 100?.

a.) How long will it take for the capacitor to release 50% of its stored energy through the flashbulb?







b.) What was the average power burned by the resistor over that time?

Explanation / Answer

Im gonna call q=4uC and Q=-9uC for throughtout this thing 1. kqQ/r^2=F r=0.3 plug in everything u get -3.569N negative means to the left. 2. E=kq/r^2-kQ/R^2 r=0.1, R=0.2 plug in everything u get 5618750. To the right since its positive. 3. kq/r^2=kQ/(r+0.3)^2 Just imague where could it be? and you would see that it should be to the left of q, since its weaker than Q, and its oppsoite charged. If you can't understand this simpel reasoning, then read over ur book again, cuz there's no point to continue on if u dont' get that. the reason u put r+0.3 is because its' 0.3 between Q and q. and r is the distance to the point (ur answer) In this case, ur assuming left is positive, which you will have to change to negative later when u get the asnwer. Plug in everything and solve for r, r=-3/25, 3/5. You want the positive answer. which is 3/5. Then you change it back to negative, cuz ur assuming q is the origin. and left is negative. so its neutral at 3/5 meters to the left of the 4uC charge. 4. V=kq/r+kQ/R r=0.1, R=0.2, cuz its 0.1 to the right of q, and 0.2 to the left of Q. plug in everything u get V=-44950. 5. W=(q-Q)V V is ur previous answer. plug in everythign u get, -0.58435J. It makes sense to be negative cuz ur putting in energy to separate them, if its positive, it means the two charges repel each other, so they put in energy on their own. 6. q=CV =1.35X30^-3=0.0405C then use this equation q=CVb(1-e^(-t/RC)) q=0.0405 C is capacitance, Vb is 1.5, R is resistance. plug in everything solve for t, u get 345.39seconds. 6b. U=1/2(CV^2) same value plug it in, u get 0.03375J U=q^2/2C then solve for q q=0.045. Which is the answer you got earlier :) 7. 50 percent of charges means 0.045 divide by 2 which is 0.0225. so use this equation cuz ur releasing charges you HAVE to use this equation instead of the earlier one. q=qo(1-e^(-t/RC)) plug in everything u get t=69.31seconds. 7b. use this equation again. q=qo(1-e^(-t/RC)) but this time, use t=69.31 divide by 2, which is 34.65 second. plug in everything and oslve for q u get a value. But i dont have time to solve now i gotta go soon. This is the average charge. q=CV, plug in everything and solve for V, C=30X10^-3 and q is ur average charge. then use this equation U=V^2/R and plug in everything, u get ur heat through.

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