Answer:
E= 4.35*10^6 N/C
Explanation:
Let's find the area charge density of the plate
α= 6.9*10^-6/9*10^-2 = 7.7*10^-5C/m2
Now we can calculate the electric field just of the plate
E =α/2e =7.7*10^-5/2*8.85*10^-12 = 4.35*10^6 N/C
Answer:
1.75atm
Explanation:
According to Boyle's law, the pressure P of a fixed mass of gas is inversely proportional to it's volume V provided that the temperature remains constant.

This implies the following;
Provided temperature is kept constant.
Given;

From equation (1), we can write;

Since all the units are consistent, there is no need for conversion.
Answer:
The resistance that will provide this potential drop is 388.89 ohms.
Explanation:
Given;
Voltage source, E = 12 V
Voltage rating of the lamp, V = 5 V
Current through the lamp, I = 18 mA
Extra voltage or potential drop, IR = E- V
IR = 12 V - 5 V = 7 V
The resistance that will provide this potential drop (7 V) is calculated as follows:
IR = V

Therefore, the resistance that will provide this potential drop is 388.89 ohms.
Any one trial might have been done incorrectly.
Answer:
a force and a movement in the same direction as the force.
Explanation: