Which of the following will increase the capacitance of a parallel-plate capacit

Question

Which of the following will increase the capacitance of a parallel-plate capacitor?
D) an increase in the plate area and a decrease in the plate separation
2) The plates of a parallel-plate capacitor are maintained with constant potential by a battery as
they are pulled apart. During this process, the amount of charge on the plates
C) must decrease.
3) A bug zapper consists of two metal plates connected to a high-voltage power supply. The
voltage between the plates is set to give an electric field slightly less than 1×10^6 V/m. When a bug
flies between the two plates, it increases the field enough to initiate a spark that incinerates the
bug. If a bug zapper has a 3000 V power supply, what is the approximate separation between the
plates?
B) 0.3 cm

4) Sharks and related fish can sense the extremely weak electric fields emitted by their prey in the
surrounding waters. These detectors, located in their noses and called ampullae, are so sensitive
they can detect a voltage gradient of 1 V across1 cm of seawater. How far apart would you have
to place the poles of a 1.5 V battery to achieve the same electric field?
D) 1.5×10^4 m
5) What is the direction of the electric field at the dot in the figure?
A) to the left
6) A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it
were accelerated instead through a potential difference of 2V0, what speed would it gain?

A) v0/2
7) What are the units of potential difference?
B) Volts (V)
8) Microbes such as bacteria have small positive charges when in solution. Public health agencies
are exploring a new way to measure the presence of small numbers of microbes in drinking water
by using electric forces to concentrate the microbes. Water is sent between the two oppositely
charged electrodes of a parallel-plate capacitor. Any microbes in the water will collect on one of
the electrodes. On which electrode will the microbes collect?
B)Negative
9) A positive charge is brought near to a dipole, as shown in the figure below. If the dipole is free
to rotate, it

C) begins to rotate in a clockwise direction.
10) A K+ ion (mass of 6.5×10^23 kg) is accelerated through a cell membrane at 1.5 ×106 m/s^2.
What is the electric field between the membrane walls?
B) 609 N/C
11) Coulomb's law describes __________.
B) the force between two point charges
12) The electric field inside a metallic conductor is _____.

B) Zero
13) A small glass bead has been charged to 1.9 nC. What is the strength of the electric field 2.0
cm from the center of the bead? (k = 1/40 = 8.99 × 109 N • m2/C2)
C) 4.3 × 104 N/C
14) Which one of the arrows shown in the figure best represents the direction of the electric field
between the two uniformly charged metal plates?
C) A
15) At the same temperature, two wires made of pure copper have different resistances. The same
voltage is applied at the ends of each wire. The wires may differ in
A) length.
B) cross-sectional area.
D) amount of electric current passing through them.

16) A copper wire is stretched so that its length increases and its diameter decreases. What is the
result?
C) The wire's resistance increases, but its resistivity stays the same.
17) A battery is connected to a resistor. Increasing the resistance of the resistor will __________.
C) decrease the current in the circuit
18) Figure below shows a side view of a wire of varying circular cross section. Rank in order the
currents flowing in the three sections.
C) I1=I2=I3

19) An electric eel (Electrophorus electricus) can produce a shock of up to 600 V and a current of
1 A for a duration of 2 ms, which is used for hunting and self-defense. To perform this feat,
approximately 80% of its body is filled with organs made up by electrocytes. These electrocytes
act as self-charging capacitors and are lined up so that a current of ions can easily flow through
them. How much charge flows through the electrocytes in that amount of time?
A) 2×10^3 C
20) A defibrillator is a device used to shock the heart back to normal beat patterns. To do this, it
discharges a 12 F capacitor through paddles placed on the skin, causing charge to flow through
the heart. Assume that the capacitor is originally charged with 4.0 kV. What is the charge initially
stored on the capacitor?
D) 4.8×10^2 C

CAN ANYONE EXPLAIN WHY THESE ANSWERS PLEASE AND THANK YOU!! :)

Explanation / Answer

1) Capacitance is C = =eo*A/d

A is the area of the plates and d is the disatmce between the plates

since C is proportional to area A and inversely proportional to d

then to increase the C,we need to increase the plate area and a decrease in the plate seperation

2) C = eo*A/d

By pulling plates apart ,capacitance C is decreased,

since C = Q/V

here potential V is constant ,so if C deceases ,then charge Q also decreases

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