Asmall glass bead has been charged to + 30 nc . part a what is the the acceleration of a proton that is 1.5 cm from the center of the bead? input positive value if the acceleration is derected toward the bead and negative if it is directed away from the bead.

The bead has a positive charge and so does the proton (+1.6*10⁻¹⁹ C), so they will repulse each other, sending the proton away from the bead, giving it a negative acceleration. For the magnitude, let's use Coulomb's Law: F = Kqq/r², where F is force, K is the electrostatic constant (9*10⁹ N*m²/C²), the q's are the charges and r is the distance between them. Plugging in values (remember that the nano- prefix corresponds to 10⁻⁹ and the centi- prefix is 10⁻²), we get F = (9*10⁹)*(30*10⁻⁹)(1.6*10⁻¹⁹)/(1.5*10⁻²)² = 1.92 *10⁻¹³ N. Ok, now that we have the force between the glass bead and the proton, we can use Newton's 2nd law: F = ma, where m is mass of the proton (1.67*10⁻²⁷ kg) and a is acceleration, to find the acceleration. Solving for a, a = F/m = (1.92 *10⁻¹³ N)/(1.67*10⁻²⁷ kg) = 1.15*10¹⁴ m/s².

Try c ? that’s what i got