Solve Electrical Problems: Airplane in Thundercloud, Point Charges & More!

[thspoq2]

Im doing some homework problems and I cannot figure out how to do these for the life of me. Any help would be appreciated.

1. An airplane is flying through a thundercould at a height of 2000 m. If there is a charge concentration of +40 C at 3000 , within the cloud and -40 C at height 1000 m, what is the electric field at the aircraft?

2. Two point charges of values +3.4 and +6.6 C are separated by .10 m. What is the electrical potential at the point midway between the two point charges?

3. If a lamp has resistance of 120 Ohms when it operates at 100W, what is the applied voltage?

4. The temperature coefficient of resistivity for a "perfect" ohmic material would be?

 

[thspoq2]

Sorry, make it 5.

5. A pair of parallel plates, forming a capacitor, are charged. The plates are pulled apart to double the original seperation, the charges on the plates remaining the same. What is the ratio of the final energy stored to the original energy stored?

 

[quicknote]

1. To solve this problem, we can use the equation E = k(Q/r^2), where E is the electric field, k is the Coulomb's constant (9x10^9 Nm^2/C^2), Q is the charge concentration, and r is the distance from the source charge. In this case, we have two point charges, one at 3000 m with a charge of +40 C and one at 1000 m with a charge of -40 C. We can calculate the electric field at the aircraft by adding the electric fields due to each point charge. The electric field due to the +40 C charge would be E = (9x10^9 Nm^2/C^2)(40 C)/(2000 m)^2 = 9000 N/C. The electric field due to the -40 C charge would be E = (9x10^9 Nm^2/C^2)(-40 C)/(1000 m)^2 = -144000 N/C. Therefore, the total electric field at the aircraft would be 9000 N/C - 144000 N/C = -135000 N/C. This means that the electric field at the aircraft is directed downward with a magnitude of 135000 N/C.

2. To find the electrical potential at the point midway between the two point charges, we can use the equation V = k(Q/r), where V is the electric potential, k is the Coulomb's constant, Q is the charge of the point charge, and r is the distance from the point charge. In this case, the point midway between the two charges would be at a distance of 0.05 m from each charge. Therefore, the electric potential at this point would be V = (9x10^9 Nm^2/C^2)((3.4 C)/(0.05 m) + (6.6 C)/(0.05 m)) = 540000 V. This means that the electric potential at the point midway between the two charges is 540000 V.

3. To find the applied voltage, we can use Ohm's Law, which states that V = IR, where V is the voltage, I is the current, and R is the resistance. In this case, we are given the resistance (120 Ohms) and the power (100W). We can use the equation P = VI, where P is power, V is