**Problem 2: Electric Charge Configuration**Three point charges are fixed in place at the corners of a trapezoid as shown. The upper side has length 3.0 cm, the base has length 5.0 cm, and the height is 3.0 cm. The charges are as follows:- $ q_1 = +10.0 \, \text{nC} $- $ q_2 = -5.0 \, \text{nC} $- $ q_3 = +5.0 \, \text{nC} $Questions:a. **Electric Field at Point P** Find the electric field at point P, the lower left corner of the trapezoid.b. **Electric Potential at Point P** Find the electric potential at point P, taking the potential at infinity to be zero.c. **Motion of a Proton** Suppose a proton ($ m = 1.67 \times 10^{-27} \, \text{kg} $) is placed at point P and is released from rest. What will its speed be when it is very far from the arrangement of charges?**Diagram Explanation:**- The diagram depicts a trapezoid with point charges at each corner.- $ q_1 $ and $ q_2 $ are located at the top corners, each 3 cm apart.- $ q_3 $ is at the bottom right corner, 1 cm from the base's end, making the base 5 cm long.- Point P is located at the lower left corner of the trapezoid, 1 cm from $ q_1 $. This problem requires understanding concepts of electric fields and potentials in the context of a multiple-charge system.

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**Problem 2: Electric Charge Configuration**

Three point charges are fixed in place at the corners of a trapezoid as shown. The upper side has length 3.0 cm, the base has length 5.0 cm, and the height is 3.0 cm. The charges are as follows:

- $ q_1 = +10.0 \, \text{nC} $
- $ q_2 = -5.0 \, \text{nC} $
- $ q_3 = +5.0 \, \text{nC} $

Questions:

a. **Electric Field at Point P**
Find the electric field at point P, the lower left corner of the trapezoid.

b. **Electric Potential at Point P**
Find the electric potential at point P, taking the potential at infinity to be zero.

c. **Motion of a Proton**
Suppose a proton ($ m = 1.67 \times 10^{-27} \, \text{kg} $) is placed at point P and is released from rest. What will its speed be when it is very far from the arrangement of charges?

**Diagram Explanation:**
- The diagram depicts a trapezoid with point charges at each corner.
- $ q_1 $ and $ q_2 $ are located at the top corners, each 3 cm apart.
- $ q_3 $ is at the bottom right corner, 1 cm from the base's end, making the base 5 cm long.
- Point P is located at the lower left corner of the trapezoid, 1 cm from $ q_1 $.

This problem requires understanding concepts of electric fields and potentials in the context of a multiple-charge system.

Expert Solution

Step 1

Given:

Length of upper side of the trapezoid $= 3.0 c m$

Length of the base of the trapezoid $= 5.0 c m$

Height of the trapezoid $= 3.0 c m$

$q_{1} = 10.0 n C, q_{2} = - 5.0 n C a n d q_{3} = + 5.0 n C$

Calculation:

(a) The electric field at point P is calculated as follows.

Let us first of all redraw the given figure in a simple way.

$N o w l e t u s c a l c u l a t e t h e d i s \tan c e A P b e t w e e n t h e c h a r g e q_{1} a n d p o i n t P w i t h t h e h e l p o f " P y t h a g o r i u m t h e o r e m " . A P = \sqrt{{(A Q)}^{2} + {(P Q)}^{2}} = \sqrt{{(3)}^{2} + {(1)}^{2}} = \sqrt{10} c m = 3.2 c m = 0.032 m S o t h e f i e l d a t " P " d u e t o q_{1} i s E_{1} = \frac{1}{4 π ε_{0}} \frac{q_{1}}{{(A P)}^{2}} = (9 \times 10^{9}) \frac{10 \times 10^{- 9}}{{(0.032)}^{2}} = 87890.6 \frac{V}{m} T h e d i r e c t i o n o f t h i s f i e l d i s " a w a y " f r o m q_{1} . N o w w e w i l l c a l c u l a t e a t P f i e l d d u e t o q_{3} . N o w d i s \tan c e C P b e t w e e n q_{3} a n d P i s 5 c m o r 0.05 m . S o f i e l d a t P d u e t o q_{3} i s, E_{3} = \frac{1}{4 π ε_{0}} \frac{q_{3}}{{(C P)}^{2}} = (9 \times 10^{9}) \frac{5 \times 10^{- 9}}{{(0.05)}^{2}} = 18000 \frac{V}{m} T h i s f i e l d i s d i r e c t e d " a w a y " f r o m q_{3} . N o w t o c a l c u l a t e r e s u l \tan t o f E_{1} a n d E_{3} w e n e e d t o f i n d t h e v a l u e o f ∠ A P Q . F r o m ∆ A P Q \tan (∠ A) = \frac{1}{3} ∴ ∠ A = \tan^{- 1} (\frac{1}{3}) = 18.43 ° H e n c e, ∠ A P Q = ∠ P = [180 ° - (90 ° + 18.43 °)] = 71.57 ° H e n c e, t h e r e s u l t a n t o f E_{1} a n d E_{3} i s E_{13} = \sqrt{E_{1}^{2} + E_{3}^{2} + 2 E_{1} E_{3} \cos (71.57 °)} = \sqrt{{(87890.6)}^{2} + {(18000)}^{2} + 2 \times (87890.6) \times (18000) \times 0.316} = 95124.13 \frac{V}{m} N o w w e w i l l c a l c u l a t e f i e l d d u e t o q_{2} . F i r s t w e w i l l c a l c u l a t e t h e d i s \tan c e B P b e t w e e n q_{2} a n d p o i n t P u \sin g " P y t h a g o r i u m t h e o r e m " . B P = \sqrt{{(B R)}^{2} + {(P R)}^{2}} = \sqrt{{(3)}^{2} + {(4)}^{2}} = 5 c m = 0.05 m . S o, t h e f i e l d d u e t o q_{2} i s E_{2} = \frac{1}{4 π ε_{0}} \frac{q_{2}}{{(B P)}^{2}} = (9 \times 10^{9}) \times \frac{5 \times 10^{- 9}}{{(0.05)}^{2}} = 18000 \frac{V}{m} B u t t h e d i r e c t i o n o f t h e f i e l d i s " t o w a r d s " q_{2} a s m a g n i t u d e o f q_{2} i s n e g a t i v e . S o t h e r e s u l \tan t e l e c t r i c f i e l d a t P t a k i n g a l l t h i s c o n t r i b u t i o n s i s E_{R} = \sqrt{E_{13}^{2} + E_{2}^{2} + 2 E_{13} E_{2} \cos (180 °)} = \sqrt{{(95124.13)}^{2} + {(18000)}^{2} + 2 \times (95124.13) \times (18000) \times (- 1)} = 77124.13 \frac{V}{m} S o t h e m a g n i t u d e o f r e s u l \tan t f i e l d a t P i s 77124.13 \frac{V}{m} T h i s f i e l d i s d i r e c t e d a w a y f r o m p o i n t P .$

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