<h2>Answer:</h2><h3>(A) the positively charged surface increases and the energy stored in the capacitor increases.</h3>
When charging a capacitor transferring charge from one surface to the other, the first surface becomes negatively charged while the second surface becomes positively charged. As you transfer the charge, the voltage of the positively charged surface increases and the energy stored in the capacitor also increases. We can solve this by the definition of <em>capacitance</em><em> </em>that is <em>a measure of the ability of a capacitor to store energy. </em>For any capacitor, the capacitance is a constant defined as:
To maintain 
constant, if Q increases V also increases.
On the other hand, the potential energy 
can be expressed as:
In conclusion, as Q increases the potential energy also increases.
Answer:
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Explanation:
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Answer:
17280 J or 17.28 kJ
Explanation:
Given that the voltage drop,
U = U2 - U1
U = 9 - 6
U = 3V
Also, we're told that the current, I is equal to 20 mA with the discharge time, t being 80 hrs.
Converting the time from h oi urs to seconds, we have
t = 80 * 3600
t = 288000
Now, to find the energy needed, we're going to use the formula
w = pt, where p = U * I
p = 3 * 20*10^-3
p = 60*10^-3
w = 60*10^-3 * 288000
w = 17280 J or 17.28 kJ
Therefore, the total energy the battery delivers in the 80 hrs is 17.28 kJ
