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cestrela7 [59]
3 years ago
9

Two point charges of equal magnitude (and opposite sign) are 7.5 cm apart. At the midpoint of the line connecting them, their co

mbined electric field has a magnitude of 45 N/C. What is the magnitude of each charge?
Physics
1 answer:
Shkiper50 [21]3 years ago
7 0

Comment

The only reason you can do this is that the charges are the same. If they were not, the problem would not be possible.

Equation

The field equation is, in its simplest form,

E = kq/r^2

So each of the charges are pulling / pushing in the same direction. The equation becomes.

kq/r^2 - (-kq/r^2) = Field magnitude in N/C

Givens

  • K = 9 * 10^9 N m^2 / c^2
  • E = 45 N/C
  • r = 7.5/2 = 3.75 cm * ( 1 m / 100 cm) = 0.0375 m
  • Find Q

Solution

k*q/0.0375 ^2 - (-kq/0.0375^2) = 45 N/C           Combine

2*k*q / 0.0375^2 = 45 N/C                                  Divide by 2

kq /(0.0375^2) = 22.5 N/C                                   Multiply by 0.0375^2

kq = 22.5 * 0.0375 ^2                                           Find d^2

kq = 22.5 * 0.001406                                            Combine

kq = 0.03164 N/C * m^2                                        Divide by k

q = 0.03164 N * m^2 /C / 9*10^9 N m^2 / c^2

q = 2.84760 * 10 ^8 C

I've left the cancellation of the units for you. Notice that only 1 C is left and it is in the numerator as it should be.


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Hint: Here, we can see that 12 and 13 are consecutive numbers. So, all numbers between squares of 12 and 13 are non-square numbers. Therefore, first find squares of 12 and 13 and then subtract square of 12 from square of 13, we get numbers of non-square numbers. At the last subtract 1 from the result obtained as both extremes numbers are not included.

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