The bearing could be the below:
oppositely charged, same initial direction
same charge, opposite initial direction
You can decide by utilizing your correct hand and put your fingers toward the attractive field (North to South). Thumb toward present or charged molecule. The course of your palm will demonstrate the heading of compelling set on a decidedly charged molecule and the bearing of the back of your hand will demonstrate the bearing of a contrarily charged molecule.
Kinetic energy and potential energy pair is the quantity in which one will increase then other will decrease
As we know that sum of kinetic energy and potential energy will always remain conserved
So here we will have

so here as we move away from mean position the kinetic energy will decrease while at the same time potential energy will increase.
So the pair of potential energy and kinetic energy will satisfy the above condition
Answer: 12,600,000Cm
Explanation:
From the data's;
Charges(q) = 1.8 PC equal to 1.8 x 10^¹²C
Distance = 7 micrometer, is equal to 0.0000070m
From the equation of electric dipole moment, p= q x d, where q= charge, d=distance and p is the dipole moment.
Then we have 1.8x10^¹² x 0.0000070= 12,600,000Cm
NB: The charges are identical.
Answer:
So the ratio will be 
Explanation:
We have given heat engine absorbs 450 joule from high temperature reservoir
So 
As the heat engine expels 290 j
So work done W = 290 J
We know that efficiency 
It is given that efficiency of the engine only 55 % of Carnot engine
So efficiency of Carnot engine 
Efficiency of Carnot engine is 

