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T w o s m a l l i n s u l a ti ng s ph e r e s eac h of m a s s m = 0 . 3 0 milli g r a m s ca r r y c h a r g e s of e qu a l m a gn it ude ( q = 15 p C ) but oppo s it e s i gn a r e c onn ec t e d by a li gh t , r i g i d, i n s u l a ti ng r od o f l e ng t h d = 5 . 0 c m , a s s ho w n a bov e . T he c h a r g e s a r e p l ace d a d i s t a n c e r = 10 c m awa y fr om a l a r ge c ondu c ti ng s h ee t ca rr y i ng a un i f o r m s u r f ac e c h a r ge d e n s it y σ = 800 n C / m 2 ( t he s h ee t i s o r i e n t e d p e r p e nd i c u l a r t o t he p a g e ; t he v e r ti ca l li ne r e p r e s e n t s a c r o s s s ec ti on ) . Y ou ca n a ss u m e t h a t t he on l y f o r ce s ac ti ng i n t h i s p r ob l e m a r e e l ec t r i ca l ( i gn o r e g r a v it a ti on a l e ff ec t s ) . ( a ) C a l c u l a t e t he e l ec t r i c f i e l d due t o t he s h ee t of c h a r ge a t t he l o ca ti on of t he d i po l e ( you ca n u s e t he r e s u l t w e d e r i v e d i n c l a ss ; no n ee d t o r e d e r i ve u s i n g Ga u ss ’ s l aw ) . ( b) S k e t c h t he e l ec t r i c f i e l d g e n e r a t e d by t he s h ee t i n t he f i g u r e a bov e . ( c ) C a l c u l a t e t he m a gn it ude of t he e l ec t r i c f i e l d d u e t o t he po s iti ve c h a r ge + q a t t he l o ca ti on o f t he n e g a ti ve c h a r ge –q. Y our a n s we r s t o p a r t s ( a ) a nd ( c ) a bove s hou l d d e m on s t r a t e t h a t t he e l ec t r i ca l i n f l u e n c e o f t he s h ee t on t he c h a r g e s w il l be m u c h l a r g e r t h a n t he e l ec t r i ca l i n f l u e n c e of t he c h a r g e s on one a no t h e r , s o f or t he r e s t of t he p r ob l e m , you ca n i gno r e t he e l ec t r i c f i e l ds a nd f o r ce s t h a t t he po i nt c h a r g e s c r ea t e . ( d) O n t he f r e e body d i a g r a m t o t he r i gh t , s k e t c h t h e f o r ce s ac ti ng on t he e l ec t r i c d i po l e . ( e ) W h a t i s t he n e t f o r c e ac ti ng on t he d i p o l e ? (f ) W h a t i s t he n e t t o r que ac ti ng on t he d i po l e a bo u t a po i nt a t t he ce n t e r of t he c onn ec ti ng r od? I nd i ca t e w h e t h e r t he t o r que i s c l o c k w i s e or c oun t e r c l o c k w i s e . A f t e r s o m e tim e , t he d i po l e r eac h e s it s e qu ili b r i um po s iti on s ho w n h e r e : ( g) W h a t i s t he n e t t o r que ac ti ng on t he d i po l e i n t h i s po s iti on? J u s ti f y your a n s we r ! ( h) A t t h i s po i n t , t he li ght r od i s r e m ov e d. A ss u mi ng t h a t t he c h a r ge –q s t a r t s fr om r e s t a d i s t a n c e 7 . 5 c m awa y f r om t he s h ee t , how f a s t w il l i t be m o v i ng w h e n i t r eac h e s t he s h ee t ?


 Two small insulating spheres each of mass m = 0.30 milligrams carry charges of equal magnitude (q =

15 pC) but opposite sign are connected by a light, rigid, insulating rod of length d = 5.0 cm, as shown

above.  The charges are placed a distance r  = 10 cm away from a large conducting sheet carrying a uniform surface charge density σ = 800 nC/m2  (the sheet is oriented perpendicular to the page; the vertical line represents a cross section).

You can assume that the only forces acting in this problem are electrical (ignore gravitational effects). (a) Calculate the electric field due to the sheet of charge at the location of the dipole (you can use the

result we derived in class; no need to rederive using Gausss law).


(b) Sketch the electric field generated by the sheet in the figure above.

 

(c) Calculate the magnitude of the electric field due to the positive charge +q at the location of the negative charge –q.


Your answers to parts (a) and (c) above should demonstrate that the electrical influence of the sheet on the charges will be much larger than the electrical influence of the charges on one another, so for the rest of the problem, you can ignore the electric fields and forces that the point charges create.

 

 

 
(d) On the free body diagram to the right, sketch the forces acting on the electric dipole.

 

(e) What is the net force acting on the dipole?



 

 

 

 

(f) What is the net torque acting on the dipole about a point at the center of the connecting rod? Indicate whether the torque is clockwise or counterclockwise.


After some time, the dipole reaches its equilibrium position shown here:         (g) What is the net torque acting on the dipole in this position?  Justify

your answer!


(h) At this point, the light rod is removed.   Assuming that the charge –q starts from rest a distance

7.5cm away from the sheet, how fast will it be moving when it reaches the sheet?


Grade:11

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