Answer:
3.86×10⁶ Newton/coulombs
Explaination:
Applying,
E = F/q....................... Equation 1
Where E = Electric Field, F = Force, q = charge.
From the question,
Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs
Substitute these values into equation 1
E = 5.4×10⁻¹/ -1.4×10⁻⁷
E = -3.86×10⁶ Newtons/coulombs
Hence the magnitude of the electric field created by the
negative test charge is 3.86×10⁶ Newton/coulombs
Answer:
83.72 K
Explanation:
= Polarizability of argon = 
I = First ionization = 1521 kJ/mol
r = Distance between atoms = 3.8 A
R = Gas constant = 8.314 J/mol K
T = Boiling point
Potential energy due to dispersion of gas is given by

Kinetic energy is given by
The potential and kinetic energy will balance each other

The boiling point of argon is 83.72 K
Answer:

Explanation:
As resistor is connected to the battery of constant EMF then the power across the resistor is given as

now if two resistors are made up of same material and of same length then due to different cross sectional area they both have different resistance
Due to different resistance they both will have different power
Since power is inversely depends on the resistance
So if the power is twice that of the other then the resistance must be half
so we have


since one resistance is half that of other resistance
So the area of one must be twice that of other
so we have



<span>Because the droplets are conductors, a droplet's positive and negative charges will separate while the droplet is in the region between the deflection plates. You are given a situation that if a neutral droplet passes between the plates. The droplet's dipole moment will point at the center of the droplet.</span>