A uniform electric field of magnitude 6.8 105 N/C points in the positive x-direction. Find the change in electric potential (?V) between the origin and each of the points given below.
(a) (0, 5.0 m)
V

(b) (5.0 m, 0)
V

(c) (5.0 m, 5.0 m)
V

Solution:

Substitute the values in the formula V=-E*d we get the answers

let us consider the electric field  that travels in a positive way or the positive direction and E denotes the magnitude

Where E= 6.8 10^5 N/C

We have the formula, V=-E*d

Take d value as 5 from the option  one

put V=0 since the fields are in the  X direction and the points are in the Y direction

By the formula,

(0, 5.0 m)

V=-E*d

V= -6.8x10^5V/m*5m

V=-3.40x10^6V

By the formula (5.0 m, 0)

V=-E*d

V= -6.8x10^5V/m*5m

V=-3.40x10^6V

By the formula ,

(5.0 m, 5.0 m)

V=-E*d

V= -6.8x10^5V/m*5m

V=-3.40x10^6V

sub V=5 we get,

V=-3.40x10^6V+5

v=2.40x10^6V

an obstruction

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.