Problem 71-12
Source: Example 16-7 - page 488 - Giancoli -
Book: Physics - 5ª edição.
Two point charges are separated by 10 cm. If Q1 = - 25 µC
and Q2 = + 50 µC, as shown by Figura 71-12.1, determine:
a) the direction and strength of the resulting electric field at the point P located between the two charges and 2 cm from the negative charge.
b) if an electron is placed at rest at point P, what will be its initial acceleration and direction?
Solution of the Problem 71-12
The resulting electric field at point P will be the sum of the fields due to the charges
Q1 and Q2. The direction of the electric field due to the
Q1 is pointing to the right, because on a negative charge the electric field points toward you. And the electric field due to Q2 also points to the right, because in a positive charge the electric field moves away from the charge. Therefore, the strength of the electric field in the
point P is given by
As r1 = 2 cm = 0,02 m, then r2 = 8 cm = 0,08 m.
Substituting for the numerical values and performing the calculation, we find
As the electron has a negative charge, it will be attracted by the positive charge and repelled by the negative charge.
Therefore, the direction of its acceleration will be to the left, in the opposite direction to the electric field. And the value of the acceleration is given by
Remembering that the mass of the electron is m = 9.1 x 10-31 Kg and making the substitution by numerical values and performing the calculation we find
