SME
Electromagnetic Theory · Problem type
The method of images: a charge above a grounded plane
Use this for “A 4 nC charge is held 2 cm above a large grounded metal plate. Find the force pulling it toward the plate, the induced charge per area just below it, and the total induced charge.”
Diagram
The grounded plane stays at V = 0: the charge and an image −q at −d below give V = 0 all over it.
F
=
kq²
÷
(2d)
2
=
8.99 ×
10
9
×
(1 ×
10
-9
)
2
÷
(2 × 0.05)
2
=
8.99 ×
10
-7
N,
toward
the
plane
Under
the
charge
σ₀
=
−
q
2πd²
=
−6.37 ×
10
-8
C/
m
2
The induced charge adds to −q.
This is an example. Type your own values to start a new problem.
F
=
k
(
q (
)
)
2
(2
×
d (
)
)
2
=
F (N)
σ₀
Induced charge per area under q
calculated
Qind
Induced charge in all
calculated
Formulas
F
=
8.99 ×
10
9
×
q
2
÷
(2 ×
d
)
2
8.99 ×
10
-7
=
8.99 ×
10
9
×
(1 ×
10
-9
)
2
÷
(2 × 0.05)
2
σ₀
=
−
q
2 × π ×
d
2
−6.3662 ×
10
-8
=
−
1 ×
10
-9
2 × π ×
0.05
2
Qind
=
−
q
−1 ×
10
-9
=
−(1
×
10
-9
)
Formulas are worked in C, m, C/m². The step-by-step shows the conversions.
Assumptions
•
The plane is a large grounded conductor, V = 0. An image charge −q at d below the plane keeps V = 0 there, so above the plane the field is that of q and −q (the uniqueness theorem).
•
The force on q is the image’s pull across 2d: F = kq² ÷ (2d)², toward the plane.
•
The induced charge per area is σ = −ε₀E at the plane; just below q it is σ₀ = −q ÷ (2πd²), and over the whole plane it adds to −q. k = 8.99 × 10⁹ N·m²/C².
Step-by-step
WE KNOW
q = 1 nC, d = 5 cm
FIND
force toward the plane (F), induced charge per area under q (σ₀), induced charge in all (Qind)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to C, m, C/m²).
q = 1 nC = 1 × 10⁻⁹ C (1 C = 10⁹ nC)
d = 5 cm = 0.05 m (1 m = 100 cm)
Step 2 · Find force toward the plane (F)
Use F = 8.99 × 10⁹ × q² ÷ (2 × d)²
Coulomb’s
law
between
q
and
its
image
−
q
,
2
d
apart:
always
a
pull
toward
the
plane.
F
=
8.99 ×
10
9
×
(
10
-9
)
2
÷
(2 × 0.05)
2
F
=
(8.99
×
10
9
)
×
(
10
-9
)
2
÷
0.1
2
F
=
(8.99
×
10
9
)
×
10
-18
0.01
F
=
8.99 ×
10
-9
0.01
F
=
8.99 ×
10
-7
N
Step 3 · Find induced charge per area under q (σ₀)
Use σ₀ = −q ÷ (2 × π × d²)
Just
below
q
,
the
field
of
q
and
its
image
is
2k
q
d
2
into
the
plane;
σ
=
−ε₀E
gives
−
q
2πd²
.
σ₀
=
−
10
-9
2 × π ×
0.05
2
σ₀
=
−1 ×
10
-9
2 × 0.0025 × π
σ₀
=
−1 ×
10
-9
0.005 × π
σ₀
=
−1 ×
10
-9
0.01571
σ₀
=
−6.3662 ×
10
-8
C/
m
2
Step 4 · Find induced charge in all (Qind)
Use Qind = −q
Every
field
line
leaving
q
ends
on
the
plane,
so
the
induced
charge
adds
to
−
q
.
Qind
=
−
10
-9
Qind
=
−1 ×
10
-9
C
Step 5 · Convert the answers
Change each answer back to the unit you chose.
σ₀ = −6.3662 × 10⁻⁸ C/m² = −63.662 nC/m² (1 C/m² = 10⁹ nC/m²)
Qind = −1 × 10⁻⁹ C = −1 nC (1 C = 10⁹ nC)
Check
Put the numbers back in. Both sides should match.
8.99 ×
10
-7
=
8.99 ×
10
9
×
(
10
-9
)
2
÷
(2 × 0.05)
2
✓
−6.3662 ×
10
-8
=
−
10
-9
2 × π ×
0.05
2
✓
−1 ×
10
-9
=
−
10
-9
✓
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