{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

646_PartUniversity Physics Solution

# 646_PartUniversity Physics Solution - Electric Charge and...

This preview shows pages 1–3. Sign up to view the full content.

Electric Charge and Electric Field 21-39 The charge is negative, so the field at the origin is directed toward the small segment. The small segment is located at angle θ as shown in the sketch. The electric field due to dQ is shown in Figure 21.97b, along with its components. 2 0 1 4 dQ dE a π = P 2 2 0 2 Q dE d a θ π = P Figure 21.97b ( ) 2 2 0 cos /2 cos x dE dE Q a d θ π θ θ = = P ( ) /2 / 2 0 2 2 2 2 2 2 0 0 0 0 cos sin 2 2 2 x x Q Q Q E dE d a a a π π θ θ θ π π π = = = = P P P ( ) 2 2 0 sin /2 sin y dE dE Q a d θ π θ θ = = P ( ) /2 / 2 0 2 2 2 2 2 2 0 0 0 0 sin cos 2 2 2 y y Q Q Q E dE d a a a π π θ θ θ π π π = = = = P P P E VALUATE : Note that , x y E E = as expected from symmetry. 21.98. I DENTIFY : Apply 0 x F = and 0 y F = to the sphere, with x horizontal and y vertical. S ET U P : The free-body diagram for the sphere is given in Figure 21.98. The electric field E G of the sheet is directed away from the sheet and has magnitude 0 2 E σ = P (Eq.21.12). E XECUTE : 0 y F = gives cos T mg α = and cos mg T α = . 0 x F = gives 0 sin 2 q T σ α = P and 0 2 sin q T σ α = P . Combining these two equations we have 0 cos 2 sin mg q σ α α = P and 0 tan 2 q mg σ α = P . Therefore, 0 arctan 2 q mg σ α = P . E VALUATE : The electric field of the sheet, and hence the force it exerts on the sphere, is independent of the distance of the sphere from the sheet. Figure 21.98 21.99. I DENTIFY : Each wire produces an electric field at P due to a finite wire. These fields add by vector addition. S ET U P : Each field has magnitude 2 2 0 1 4 Q x x a π + P . The field due to the negative wire points to the left, while the field due to the positive wire points downward, making the two fields perpendicular to each other and of equal magnitude. The net field is the vector sum of these two, which is E net = 2 E 1 cos 45° = 2 2 0 1 2 cos45 4 Q x x a π ° + P . In part (b), the electrical force on an electron at P is eE . E XECUTE : (a) The net field is E net = 2 2 0 1 2 cos45 4 Q x x a π ° + P . E net = ( )( ) 9 2 2 6 2 2 2 9.00 10 N m /C 2.50 10 C cos45 (0.600 m) (0.600 m) (0.600 m) × × ° + = 6.25 × 10 4 N/C. The direction is 225° counterclockwise from an axis pointing to the right through the positive wire.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document