Explanation:
Using Ohm's Law and a bit of substitution, we can use voltage divided by current to solve for resistance. Doing that, we'll get 6 Ohm.
Answer:
The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.
F = N e E where E is the value of the field and N e the charge Q
M g = N e E and M g is the weight of the drop
N = M g / (e E)
N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13
.00011 kg is a very large drop
Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs
Check: N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13 electrons
Answer: If x + y = a, xxy = b and x • a = 1 , then 2 (a~ - l)a- a x b (b2 ... xy-plane, then the vector in the same plane having projections
Answer:
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.
Explanation:
From coulomb's law, F = Eq
Thus,
F = E₁q₁
F = E₂q₂
Then
E₂q₂ = E₁q₁

where;
E₂ is the external electric field due to second test charge = ?
E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C
q₁ is the first test charge = 13 mC
q₂ is the second test charge = 23 mC
Substitute in these values in the equation above and calculate E₂.

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.
However, the direction of the external field is still to the right.
The answer yr looking for would b true!