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Lesson_2.3_Homework_and_Solutions

# Lesson_2.3_Homework_and_Solutions - Lesson 2.3 Homework...

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Lesson 2.3 Homework Solutions 1. Field along the axis of a charged disc is: 2 2 2 2 2 2 2 2 3/ 2 2 2 3/ 2 2 1 2 2 ( ) ( ) x E k x R dE x R x k R k dx x R x R π σ π σ π σ = - + + - = - = + + This expression cannot be zero for any value of x. There are no point along the axis where a maximum occurs. The maximum value of the field is at the surface of the plane and is equal to 2 π k σ . 2 π k σ E x x 2. If σ is the surface charge density of the glass pane, the electric field near it E = 2 π k σ . Since the Styrofoam ball is placed in this field, it experiences a force F = Eq = 2 π k σ q When the ball floats motionless, this force is balanced by the weight of the ball. 2 π k σ = mg 3 7 2 9 6 2.5 10 9.8 2 2 9 10 2.8 10 1.55 10 mg kq C m σ π π - - - - × × = = = × × × × ×

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3. (i) If σ is the surface charge density on the plates, each plate will produce a field 2 π k σ near it. Since the field due to oppositely charged plates are in the same direction, the total field at any point between the plates is; E = 2(2 π k σ ) = 4 π k σ . 2(2 π k σ29 σ29σ29 6 5 2 6 6 9 4 3.0 10 3.0 10 3.0 10 4 4 9 10 2.65 10 C k k m π σ σ π π - - = × × × = = = × × × (ii) total charge per square centimeter = 5 5 2 4 2.65 10 2.65 10 10 C cm - - - × = × If n is the number of fundamental charges, ne = 2.65 × 10 -9 . But e = 1.602
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