Charge on can A is positive.
Charge on can C is negative.
Punctuation and capitalization are very useful things to pay attention to and this question would be a lot easier to understand if you had actually used both capitalization and punctuation. If I'm understanding the question, you have 3 metal can that are insulated from the environment and initially touching each other in a straight line. Then a negatively charged balloon is brought near, but not touching one of the cans in that line of cans. While the balloon is near, the middle can is removed. Then you want to know the charge on the can that was nearest the balloon and the charge on the can that was furthermost from the balloon.
As the balloon is brought near to can a, the negative charge on the balloon repels some of the electrons from can a (like charges repel). Some of those electrons will flow to can b and in turn flow to can c. Basically you'll have a charge gradient that's most positive on that part of the can that's closest to the balloon, and most negative on the part of the cans that's furthest from the balloon. You then remove can B which causes cans A and C to be electrically isolated from each other and prevents the flow of elections to equalize the charges on cans A and C when the balloon is removed. So you're left with a deficiency of electrons on can A, so can A will have a positive overall charge, and an excess of electrons on can C, so can C will have a negative overall charge.
Answer: this would be D.) What does a turntable built into a track look like
Explanation:
Heat = mass (m)*specific heat (C)* change in temperature (Δt)
In the current scenario,
mass = 200 g = 0.2 kg
C = 0.11 kCal/kg.°C
Δt = 10 °C
Therefore,
Heat = 0.2*0.11*10 = 0.22 kCal = 0.22*4186 J = 920.92 J
Answer:
(a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
Explanation:
Given that,
Charge = 10.1 μC
Capacitor C₁ = 1.10 μF
Capacitor C₂ = 1.92 μF
Capacitor C₃ = 1.10 μF
Potential V₁ = 51.5 V
Let V₁ and V₂ be the potentials on the two plates of the capacitor.
(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor
Using formula of potential difference
Put the value into the formula
The potential on the second plate
(b). We need to calculate the equivalent capacitance of the two capacitors
Using formula of equivalent capacitance
Put the value into the formula
Hence, (a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
