A +13.4 nC charge is located at (0,9.4) cm and a -4.23 nC charge is located (4.99, 0) cm. Where would a -14.23 nC charge need to
1 answer:
In order to solve this problem, we will first need to find the electric field at the origin without the 3rd charge
E1 = (9x10^9)(13.4x10^-9)/(9.4x10^-2)^2 = 13648.7 V/m towards the negative y-axis
E2 = (9x10^9)(4.23x10^-9)/(4.99x10^-2)^2 = 15289.1 V/m towards the positive x-axis
The red arrow shows the direction of which the electric field points.
To make the electric field at the origin 0, we must find a location where q3 = the magnitude of q1 and q2
Etotal = sqrt(E1+E2) = 20494.97 V/m
E3 = 20494.97 = (9x10^9)(14.23x10^-9)/(d)^2
d = 0.079 m = 7.9 cm
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<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.