Two point charges&nbsp;q1=+2.40&nbsp;nC and&nbsp;q2=−6.50&nbsp;nC are 0.100 m apart. Point A is midway between them and point B is 0.080 m from&nbsp;q1&nbsp;and 0.060 m from&nbsp;q2. Take the electric potential to be zero at infinity.(a) Find the potential at point A.(b) Find the potential at point B.(c) Find the work done by the electric field on a charge of 2.50 nC that travels from point B to point A.

VA=14πϵ0[q1r1+q2r2] =8.99×109[2.4×10−9.08−6.5×10−9.05] =−737.18 V =8.99×109[2.4×10−9.08−6.5×10−9.06] =−704.18 V VB−VA=−704.1867−(−737.18) =32.9933 V work done = qV W=2.5×10−9×32.9933=82.48325 J

PART A) at point A electric potential V = (kq1/r) + (kq2/r) where r=0.050m,q1=2.40nC,q2=−6.40nC,k=8.98⋅109NC2m2 ⇒V=(8.98∗⋅109)(2.40−6.40)⋅10−90.05=−718.4v PART B) at point B r1=0.08m,r2=0.06m V=(kq1r1)+(kq2r2)=(8.98⋅109)⋅[(2.40⋅10−90.08)−(6.40⋅10−90.06)]=−688.46v PART C) potential difference between A and B = -718.4 -(-688.46) = - 29.94 v workdone = change in potential energy = q⋅δV=2.60⋅10−9⋅(−29.94)=−77.84nJ

VA=14πϵ0[q1r1+q2r2] =8.99×109[2.4×10−9.08−6.5×10−9.05] =−737.18 V =8.99×109[2.4×10−9.08−6.5×10−9.06] =−704.18 V VB−VA=−704.1867−(−737.18) =32.9933 V work done = qV W=2.5×10−9×32.9933=82.48325 J

PART A) at point A electric potential V = (kq1/r) + (kq2/r) where r=0.050m,q1=2.40nC,q2=−6.40nC,k=8.98⋅109NC2m2 ⇒V=(8.98∗⋅109)(2.40−6.40)⋅10−90.05=−718.4v PART B) at point B r1=0.08m,r2=0.06m V=(kq1r1)+(kq2r2)=(8.98⋅109)⋅[(2.40⋅10−90.08)−(6.40⋅10−90.06)]=−688.46v PART C) potential difference between A and B = -718.4 -(-688.46) = - 29.94 v workdone = change in potential energy = q⋅δV=2.60⋅10−9⋅(−29.94)=−77.84nJ

"Two point chargesq1=+2.40nC andq2=−6.50nC are 0.100 m apart...." solved and explained

Amari Flowers

Answered question

2021-03-02

Two point charges $q_{1} = + 2.40$ nC and $q_{2} = - 6.50$ nC are 0.100 m apart. Point A is midway between them and point B is 0.080 m from $q 1$ and 0.060 m from $q_{2}$ . Take the electric potential to be zero at infinity.

(a) Find the potential at point A.
(b) Find the potential at point B.
(c) Find the work done by the electric field on a charge of 2.50 nC that travels from point B to point A.

Answer & Explanation

broliY

Skilled2021-03-03Added 97 answers

V_{A} = \frac{1}{4 π ϵ_{0}} [\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}}]

= 8.99 \times 10^{9} [\frac{2.4 \times 10^{- 9}}{.08} - \frac{6.5 \times 10^{- 9}}{.05}]

= - 737.18 V

= 8.99 \times 10^{9} [\frac{2.4 \times 10^{- 9}}{.08} - \frac{6.5 \times 10^{- 9}}{.06}]

= - 704.18 V

V_{B} - V_{A} = - 704.1867 - (- 737.18)

= 32.9933 V

work done = qV

W = 2.5 \times 10^{- 9} \times 32.9933 = 82.48325 J

Jeffrey Jordon

Expert2021-09-30Added 2605 answers

PART A) at point A

electric potential V = (kq1/r) + (kq2/r)

where $r = 0.050 m, q_{1} = 2.40 n C, q_{2} = - 6.40 n C, k = 8.98 \cdot 10^{9} N \frac{C^{2}}{m^{2}}$

$\Rightarrow V = (8.98 * \cdot 10^{9}) (2.40 - 6.40) \cdot \frac{10^{-} 9}{0.05} = - 718.4 v$

PART B) at point B

$r_{1} = 0.08 m, r_{2} = 0.06 m$

$V = (\frac{k q_{1}}{r_{1}}) + (\frac{k q_{2}}{r_{2}}) = (8.98 \cdot 10^{9}) \cdot [(2.40 \cdot \frac{10^{-} 9}{0.08}) - (6.40 \cdot \frac{10^{-} 9}{0.06})] = - 688.46 v$

PART C) potential difference between A and B = -718.4 -(-688.46) = - 29.94 v

workdone = change in potential energy = $q \cdot δ V = 2.60 \cdot 10^{- 9} \cdot (- 29.94) = - 77.84 n J$

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