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natita [175]
3 years ago
9

A Capacitor is a circuit component that stores energy and can be charged when current flows through it. A current of 3A flows th

rough a capacitor that has an initial charge of 2μC (micro Coulombs). After two microseconds, how much is the magnitude of the net electric charge (in μC) of the capacitor?
Physics
1 answer:
ddd [48]3 years ago
4 0

Answer:

8\mu C

Explanation:

t = Time taken = 2\mu s

i = Current = 3 A

q(0) = Initial charge = 2\mu C

Charge is given by

q=\int_0^t idt+q(0)\\\Rightarrow q=\int_0^{2\mu s} 3dt+2\mu C\\\Rightarrow q=3(2\mu s-0)+2\mu C\\\Rightarrow q=6\mu C+2\mu C\\\Rightarrow q=8\mu C

The magnitude of the net electric charge of the capacitor is 8\mu C

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A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric
Liula [17]

The electric field produced by a single-point charge is given by

E(r)=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge


To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.


1) The first charge is q=-3.00 nC=-3.00 \cdot 10^{-9} C, and it is located at x=0, so its distance from the point x=0.200 m is

r=0.200 m-0=0.2 m

Therefore, the electric field is

E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.


2) The second charge is q=-5.50 nC=-5.5 \cdot 10^{-9}C and it is located at x=0.800 m, so its distance from the point is

r=0.800 m-0.200 m=0.6 m

Therefore, the electric field is

E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.


3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C

and the sign tells us that the field is directed toward negative x-direction.

7 0
3 years ago
Using a complete sentence state what would most likely happen to the production of oxygen by duckweed plans if the intensity and
jeka94

Answer:

This question will be answered based on general photosynthetic understanding. The answer is:

The production of oxygen would increase

Explanation:

The characteristics of most plant forms is their ability to photosynthesize i.e. use solar energy (from sunlight) to make food (chemical energy). The product of this photosynthetic process is OXYGEN gas, which is released as a waste product via the stomata on their leaves. Note that, photosynthesis cannot occur without LIGHT as it provides the energy needed for the process.

Hence, in the duckweed plant like every other photosynthetic plant, the increase in the intensity and duration of exposure to light means the rate at which photosynthesis occurs will be increased. An increased photosynthetic rate means the synthesis of the products will also be increased i.e. glucose and OXYGEN.

6 0
3 years ago
If a 2 x 10^-4C test charge is given 6.5J of energy, determine the electric potential difference.
Gwar [14]

Answer:

The electric potential difference is 32500 volt.

Explanation:

Given that,

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Using formula of potential difference

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