Aconstant electric field of magnitude e = 148 v/m points in the positive x-direction. how much work (in j) does it take to move the charge q =-13 μc from x1 = -81 m, y1 = -131 m, to x2 = 107 m, y2 = 76 m? ( note: the charge is given in micro-coulomb. the symbol in front of the c should be a greek letter mu.)

Explanation:

Given:

electric field,

charge,

initial position coordinates,

final position coordinates,  

We find the distance through which the charge has been moved:

Where we have (x1,y1) & (x2,y2) as the initial and final coordinates of the points.

Now we need the angle through which displacement is made with respect to the direction of electric field.

Now from the relation between the change in potential difference:

∵The change in voltage is defined as the work done per unit charge.

∴

Putting the respective values

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y