4. A positively charged particle of mass m = 9.0 g and Q-2.6 × 10-6 C moves para

Question

4. A positively charged particle of mass m = 9.0 g and Q-2.6 × 10-6 C moves parallel to the ground in the positive x direction with acceleration of magnitude 4.0 m/s2. The charge is not touching the ground and is in a uniform electric field. Assume that the only forces acting are the electrostatic force and the gravitational force. ty +X ground a) Find the magnitude and direction of the vertical component of the electric field. (3.5) b) Find the magnitude and direction of the horizontal component of the electric field. (3.5)

Explanation / Answer

since the charge is moving in horizontal direction, the force acting on the charge is electric force due to electric fieldalong x aixs

(a)

electric field along vertical direction is zero

(b)

Apply Newton second law

F =ma

Eq = ma

E = ma/q

= 9* 10^-3(4)/ (2.6* 10^-6)

=13.84 * 10^3 N/C

(c)

magnitude of electric field is

E = 13.84 * 10^3 N/C  

direction is along x axis

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