The x component of a uniform electric field is ${\mathrm{E}}_{\mathrm{x}}=12\mathrm{V}/\mathrm{m}$. What is the potential difference between ${\mathrm{x}}_{\mathrm{i}}=45\mathrm{m}$ and ${\mathrm{x}}_{\mathrm{f}}=12\mathrm{m}$?

An arrangement of source charges produces the electric potential V=5000x^2 along the x-axis, where V is in volts and x is in meters. What is the maximum speed of a 1.0 g, 10 nC charged particle that moves in this potential with turning points at ±8.0 cm?

The electric potential in a region of space is given by V=V₀[(x²+2y²)/(0.10 m)²], where V₀ is a constant. A proton released from rest at (x, y)=(20 cm, 0 cm) reaches the origin with a speed of 7.5×10^5 m/s .b. At what value of y on the y-axis should a He+ ion (charge +e, mass 4 u) be released from rest to reach the origin with the same speed?

INT CALC Two positive point charges q are located on the y-axis at y = ±a.e. Your answer to part d shows that an electron experiences a linear restoring force, so it will undergo simple harmonic motion. What is the oscillation frequency in GHz for an electron moving between two 1.0 nC charges separated by 2.0 mm?

The electric potential in a region of uniform electric field is $1200\mathrm{V}$ at $\mathrm{x}=-4.3\mathrm{m}$ and $2800\mathrm{V}$ at $\mathrm{x}=2.1\mathrm{m}$. What is ${\mathrm{E}}_{\mathrm{x}}$?

What is the speed of a proton that has been accelerated through $2300\mathrm{V}$?

Point A is at $-25\mathrm{V}$ and point B is at $400\mathrm{V}$. A proton moves past point A at $3.5\times {10}^5\mathrm{m}/\mathrm{s}$, what is its speed when it reaches point B?

The water molecule is a permanent electric dipole with a dipole moment of $6.2\times {10}^{-30}\mathrm{C}\mathrm{m}$. A water molecule is aligned with an electric field of magnitude $7600\mathrm{V}/\mathrm{m}$. How much energy is required to rotate the molecule 180°?