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.A 0.170 g glass bead is charged by the removal of 1.20 x10 ^10 electrons.
\What is the direction of the electric field EE⃗ at which the bead will hang suspended in the air?
\Step 1:
\The mass of the glass bead is .
The glass bead is chrged by removal of ![\"\"](\"/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=26ec1acba0a2e441312b5f95a2288e3b.png\")
electrons.
The electric field is that of a positive chrge on the glass bead.
\The charge is assumed to be a point chrge.
\The electric field is ![\"\"](\"http://www.mathskey.com/test/fckeditor/editor/plugins/fckeditor_wiris/integration/showimage.php?formula=2651ab044501d4a9372a4f3a7431b2d7.png\")
.
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.
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The direction of the electric field is away from the bead.
\The direction of the electric field is upwards.
\The glaa bead hangs suspended in the air whwn the net force acting on the bead is zero.
\The two forces that act on the bead are electric force and the weight.
\Because the bead is positively chrged , the electric field must be pointed upwards to cause an downward force, which will balance the weight.
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