The electrical force experienced by a -3.4μC charge at some point P has a magnitude of 19.8N and points due North.

Question

The electrical force experienced by a -3.4μC charge at some point P has a magnitude of 19.8N and points due North. What is the ratio of the magnitude of the acceleration of an electron to that of a proton, when the electric field acting on each has the same strength?

ae/ap = ?

Answer 1

All of the data given in the question is irrelevant. A proton and an electron experience the same magnitude of electrical force given an E-field (just in opposite directions)

The mechanics of how they will accelerate is a_i = F/m_i where i is p or e.

Therefore a_e/a_p = m_p/m_e (by taking the ratio of a's with F cancelling or realizing that a and m are inversely related if F is constant). So the electron accelerates 1836 times faster than a proton. You can find the two masses or the ratio by googling or looking in text table.

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