Two point charges, q1 -25 µc and q2 +50 µc are separated by a distance of d = 12 cm. the electric field at the point p is zero. how far from q1 is p?

Solution:  

Q1 = -27 × 10^-6 C  

Q2 = 50 × 10^-6 C  

d = 13cm = 0.13 m  

Let x be the distance between P and Q1. At point P, the field due to q1 ie. E1 and q2 ie. E2 are equal in magnitude but opposite in direction and hence cancel each other resulting in zero net electric field.  

At P  

|E1|=|E2|  

kq1/x^2 = kq2/(x+d)^2  

q1 (x^2+0.0169+0.26x)= q2 × x^2  

27x^2 + 0.4563 + 7.02x = 50x^2  

23x^2 -7.02x -0.4563 =0  

x = 0.36m = 36cm  

Or x= -0.055m= -5.5 cm  

Note: -5.5 cm corresponds to a position of P between q1 and q2 where the net electric field is also 0.

! 55 !  

write a paragraph or story to explain the transformation between potential and kinetic energy as shown in your picture or model.  

you should have at least 5 complete sentences with a main idea and supporting details.  

you must include the following words:  

potential energy

kinetic energy

mechanical energy

conservation of energy

dude, this is hard lol : ))