Question

One type of ink-jet printer, called an electrostatic ink-jet printer, forms the letters by using deflecting electrodes to steer charged ink drops up and down vertically as the ink jet sweeps horizontally across the page. The ink jet forms 30-μm-diameter drops of ink, charges them by spraying 800,000 electrons on the surface, and shoots them toward the page at a speed of 20 m/s . Along the way, the drops pass through two horizontal, parallel electrodes that are 6.0 mm long, 4.0 mm wide, and spaced 1.0 mm apart. The distance from the center of the electrodes to the paper is 2.0 cm. To form the tallest letters, which have a height of 6.0 mm, the drops need to be deflected upward (or downward) by 3.0 mm. What electric field strength is needed between the electrodes to achieve this deflection? Ink, which consists of dye particles suspended in alcohol, has a density of 800 kg/m3.

Accepted Answer

Step 1: Calculate the mass of a single ink drop. The volume of a spherical ink drop can be calculated using the formula for the volume of a sphere: $$ V = \frac{4}{3} \pi r^3 $$, where $$ r  is $$the radius of the drop. The radius is half the diameter, so $$ r = 15 	imes 10^{-6} \, 	ext{m} . $$Multiply the volume by the density of the ink ($$ ho = 800 \; 	ext{kg/m}^3 ) to $$find the mass: $$ m = ho V . $$Step 2: Determine the charge on the ink drop. The charge $$ q  is $$given by the number of electrons sprayed onto the drop multiplied by the charge of a single electron: $$ q = n e $$, where $$ n = 800,000 $$ and $$ e = 1.6 	imes 10^{-19} \, 	ext{C} . $$Step 3: Relate the electric force to the deflection. The electric force $$ F = qE  ($$where $$ E  is $$the electric field strength) causes the ink drop to accelerate vertically. Use Newton's second law $$ F = ma  to $$find the vertical acceleration $$ a $$, where $$ m  is $$the mass of the drop calculated in Step 1. Step 4: Use kinematics to find the required acceleration. The vertical deflection $$ y  is 3.0 mm$$ ($$ 3.0 	imes 10^{-3} \; 	ext{m} $$), and the horizontal distance traveled through the electrodes is 6.0 mm ($$ 6.0 	imes 10^{-3} \; 	ext{m} ). $$The time $$ t  it $$takes to traverse the electrodes is $$ t = \frac{d}{v} $$, where $$ d  is $$the length of the electrodes and $$ v = 20 \; 	ext{m/s}  is $$the horizontal velocity. Use the kinematic equation $$ y = \frac{1}{2} a t^2  to $$solve for $$ a . $$Step 5: Solve for the electric field strength $$ E . $$Rearrange the equation $$ F = qE  to $$find $$ E = \frac{F}{q} $$, where $$ F = ma $$ and $$ a  is $$the acceleration found in Step 4. Substitute the values of $$ m $$, $$ a $$, and $$ q  to $$calculate $$ E .$$

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