Along the -axis lies a region in which the electric potential varies with position as…

PHYSICS · Multi-part · CBSE · Physics · Electrostatic Potential and Capacitance · Electrostatic Potential and Capacitance

Problem

Along the

x

-axis lies a region in which the electric potential varies with position as follows: • From

x=0

x=0.050\,\text{m}

, the potential increases linearly from

0\,\text{V}

+400\,\text{V}

. • From

x=0.050\,\text{m}

x=0.100\,\text{m}

, the potential decreases linearly back to

0\,\text{V}

. Outside this region the potential is constant at

0\,\text{V}

. A proton is launched at

x=0

with initial speed

v_0

directed toward increasing

x

. Neglect all forces except the electrostatic force. (a) Determine the magnitude and direction of the electric field in each of the two segments

0\le x\le0.050\,\text{m}

and

0.050\,\text{m}\le x\le0.100\,\text{m}

. Explain your reasoning. (b) Derive an expression for the proton’s speed

v(x)

as a function of position

x

within the region

0\le x\le0.100\,\text{m}

in terms of

v_0

, the proton charge

e

, mass

m_p

, and the given potential values. (c) Using your expression from part (b), obtain an algebraic expression for the minimum initial speed

v_{0,\text{min}}

that allows the proton to reach

x=0.100\,\text{m}

. Simplify the expression as much as possible without substituting numerical values. (d) Interpret physically the significance of the quantity

v_{0,\text{min}}

you found in part (c). In particular, explain what would happen to a proton launched with (i)

v_0>v_{0,\text{min}}

and (ii)

v_0<v_{0,\text{min}}

. Justify your answers using energy considerations.