Answer:
Amplitude = 8 Volts
Frequency = 0.067 kHz
Explanation:
Note: The missing picture in question is attached for your review.
Given:
Volts/Div = 2 V/div
Time/Div = 5 msec/div
Finding Amplitude:
Now, as you can see in the attached picture, there are 4 division between two peaks of the waveform, so,
(Multiplying by 2 V/div because oscilloscope dial is set at 2 V/div)
Finding Frequency:
As can be seen in attached picture, 3 division are there for one complete cycle of waveform,so,
Since,
<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:
q=1.7346×10⁻⁶C
Explanation:
Since the electric field is perpendicular to the bottom and top of the cube,the total flux is equals the flux over the top of surface plus the flex over the lower surface
Ф(total)=Ф₃₀₀+Ф₂₃₀
But the flux is given by Ф=E.A=EACos(θ) where θ is the angle between Area vector and electric field
So
Ф(total)=E₃₀₀A Cos(180)+E₂₃₀ACos(0)
Ф(total)=A(E₃₀₀ - E₂₃₀)
The total flux is given by Gauss Law as:
Ф(total)=q/ε₀
q=ε₀Ф(total)
q=ε₀(A(E₃₀₀ - E₂₃₀))
Substitute the given values
q=(8.85×10⁻¹²){(70²)(100 - 60)}
q=1.7346×10⁻⁶C
Answer:
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Explanation: