Two small metal spheres are fixed to insulated stands and given static charges of -4.00*10^-6 C and +2.00*10^-6 C, respectively. The spheres are then placed 0.500m apart. Point P is halfway between...
Two small metal spheres are fixed to insulated stands and given static charges of -4.00*10^-6 C and +2.00*10^-6 C, respectively. The spheres are then placed 0.500m apart. Point P is halfway between the charged spheres.
a. At point P, what is the electric field strength caused by the two charged spheres(magnitude and direction)
B. waht force would an alpha particle placed at point P experience?
The arrangement of the two given charges is shown in the figure below. The electric field lines emerges from positive charges and enters the negative charges. The magnitude of the field at half distance `R =0.25 m` from each of charges is (the two fields are summing since they have the same direction at point P, towards the negative charge):
`E(R) = k_e*Q_1/R^2 +k_e*Q_2/R^2 =9*10^9*((2*10^-6)/0.25^2 +(4*10^-6)/0.25^2) =8.64*10^5 V/m`
An alpha particle is just a He nucleus (2 protons and 2 neutrons) and hence it has the electrical charge `q =+2*e` .
The force that this alpha particle experience at point P is
`F(R) = q*E(R) =2*1.6*10^-19*8.64*10^5 =2.7648*10^-13 N`
The force is directed towards the negative charge, having the same direction as the total field.